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Brief  Course  in-  Algebra 


BY 


RAYMOND  E.  MANCHESTER 

Professor  of  Mathematics.  State  Normal  School.  Oshkosh,  Wis. 
Author  of  the  Teaching  of  Mathematics 


SYRACUSE,    N.    Y. 

C.  W.  BARDEEN,    PUBLISHER 


Copyright,  1915,  by  C.  W.  Bardeen 


ERRATA 

P.     41  10th  line  for  3  read  2 
19th  line  for  3  read  2 
81   12th  line  for   -x2  read  +x2 
133   last   line  for   —  J  read   =| 

EDUCATION  DEPT. 


OUC 


TO 

/Rang  Sane  /fcancbester. 


54  1264 


During  the  past  year  or  two  there  has  been  a  grow- 
ing demand  for  a  twenty  weeks'  course  in  Algebra 
which  will  not  only  serve  as  a  preparatory  course  for 
more  advanced  mathematics,  but  will  have  an  indi- 
vidual unity  as  well.  It  must  offer  an  opportunity 
to  the  short  term  student  to  get  a  well  established 
notion  of  the  subject  through  quadratic  equations. 

This  text  is  arranged  to  satisfy  such  a  course.  All 
unnecessary  theorems,  proofs,  and  processes  have 
been  omitted.  The  problem  lists  are  very  short 
and  are  to  be  supplemented  by  the  instructor  from 
the  experience  of  the  class.  It  is,  in  fact,  a  course 
in  generalized  arithmetic,  arranged  to  face  the  de- 
mands made  by  the  ninety  per  cent  of  students  who 
expect  to  leave  school  either  during  or  at  the  expira- 
tion of  the  high  school  course. 

RAYMOND  E.  MANCHESTER 
Oshkosh,  Wis. 

June  1,  1915 


SECTION  I 

Treating  of  the  relationship  existing  between 
Algebra  and  Arithmetic. 


SECTION  II 

The  Fundamental  Operations. 

SECTION  III 
Fractions. 

SECTION  IV 
Powers  and  Roots. 

SECTION  V 

Quadratic  Equations. 


Brief  Course  in  Algebra 

SECTION  I 

Treating  of  the  relationship  existing  between  Al- 
gebra and  Arithmetic. 

A.  Letter  Symbols 

B.  Negative  Numbers 

C.  The  use  of  the  equation 

Lesson  1 

Algebra  is  an  enlarged  and  continued  course  in 
arithmetic.  As  such  it  has  a  definite  connection  with 
the  grade  work  in  arithmetic  and  such  elementary 
work  in  generalization  as  is  attempted  in  the  last 
year  of  grammar  school. 

The  enlargement  is  (1)  an  addition  to  the  Arabic 
symbol  system  by  letters  of  the  alphabet,  (to  stand 
for  undetermined  numbers);  (2)  an  increase  in  the 
usefulness  of  the  symbol  system  by  using  the  plus 
(  +  )  and  minus  (  —  )  signs  to  qualify  numbers  (as  well 
as  for  signs  of  operation) ;  and  (3)  the  use  of  the  equa- 
tion as  a  means  of  solving  problems  (made  possible 
by  the  letter  symbols).  The  study  of  arithmetical 
operations  is  continued  with  such  revisement  to  the 
rules  and  laws  as  is  necessary  to  cover  the  enlarged 
symbol  system. 

LETTER  SYMBOLS 

(1)  Throughout  the  work  in  arithmetic,  number 
ideas  have  been  expressed  by  either  a  word  (three), 

(9) 


10  BRIE?  COURSE  IN  ALGEBRA 

a  symbol  (3);  or  a  picture  (111).  But  the  word,  the 
symbol,  or  the  picture  has  stood  for  a  definite  number, 
so  that  the  student,  upon  hearing  the  word,  seeing 
the  symbol,  or  seeing  the  picture,  has  had  a  distinct 
group  brought  to  the  attention.  These  names  and 
symbols  have  been  memorized  with  the  connections 
always  made  clear. 

(2)  In  any  problem,  however,  there  is  a  certain 
number  (or  there  are  in  some  cases  certain  numbers) 
not  definitely  known.  There  being  no  definite  num- 
ber idea,  no  name,  Arabic  symbol,  or  picture  can 
stand  for  it.  We  know  that  the  number,  whatever 
it  may  be,  exists,  and  we  hope  to  determine  its  value 
by  the  solution  of  the  problem.  We  speak  of  this 
number  in  various  ways,  such  as,  "the  answer", 
"the  unknown  part",  "the  number  to  be  found",  or 
"the  result". 

Example — In  a  basket  there  are  three  times  as 
many  pears  as  apples.  The  total  number  of  both  is 
28.     To  find  the  number  of  apples. 

In  this  problem  the  unknown  is  the  number  of 
apples.  We  know  that  there  is  a  certain  number  of 
apples,  but  cannot  set  down  a  symbol  for  the  number 
until  the  problem  has  been  solved. 

Example — A  certain  number  added  to  its  half 
and  its  double  gives  a  sum  of  21.     Find  the  number. 

Here  the  unknown  number  is  the  number  itself. 
We  cannot  put  down  a  symbol  to  stand  for  it  until 
the  problem  has  been  solved. 

(3)  In  our  course  in  algebra,  we  do  have  a  symbol 
for   such   unknown   numbers.     It   being   impossible 


LETTER  SYMBOLS  11 

to  write  down  any  one  of  the  Arabic  symbols  until 
its  value  has  been  determined,  it  is  necessary  to  use 
other  symbols.  Because  of  the  familiarity  of  all 
students  with  the  letters  of  the  alphabet,  it  has  be- 
come a  custom  to  use  one  of  these  letters  as  a  symbol 
until  enough  is  known  about  the  number  to  make  it 
possible  to  use  one  of  the  Arabic  symbols. 
Example — 
In  the  first  problem  mentioned  above  we  might 

Let  #  =  the  number  of  apples. 

Then  3x  =  the  number  of  pears. 

Adding,  4#  =  the  number  of  apples  and  pears. 

Or,    4x  =  28. 

Or,  x  =  7,  the  number  of  apples. 
At  the  outset,  we  did  not  know  that  there  were  7  ap- 
ples, but  we  did  know  that  there  was  a  certain  number 
of  apples.  We  could  not  have  used  the  symbol  (3) 
or  (5)  or  (9)  or  any  of  the  other  Arabic  symbols  with 
the  certainty  that  the  statement  would  be  correct, 
so  we  allowed  (x)  to  stand  for  the  number,  whatever 
it  might  be. 

So  it  is  that  we  may  use  the  letters  to  stand  for 
numbers  until  the  numbers  have  been  so  established 
as  to  permit  us  to  use  the  Arabic  symbols. 

(4)  It  follows  that  we  may  speak  of  (a)  houses, 
(x)  horses,  (y)  trees,  or  (b)  books  when  there  is  no 
definite  number  idea  in  our  minds.  We  know  that 
there  is  a  certain  number  of  houses,  or  horses,  or 
trees,  or  books,  but  we  do  not  know  what  it  is. 
There  may  be  conditions  given  which  will  enable  us 
to  determine  the  number,  and  until  such  a  determina- 
tion is  made  the  letters  may  be  used. 


12  BRIEF  COURSE  IN  ALGEBRA 

The  advantage  of  having  such  an  addition  to  our 
symbol  system  is  at  once  apparent  when  the  solution 
of  a  difficult  problem  is  attempted. 

Example — A  young  man  wishes  to  build  a  bookcase 
having  3  shelves  arranged  so  that  the  lower  one  may 
contain  larger  books  than  the  other  two.  Suppose 
he  wishes  to  decrease  each  shelf  space  by  two  inches 
from  the  lower  to  the  upper  and  still  keep  the  book- 
case in  proportion.  Given  the  inside  measurement 
as  32  inches,  to  find  the  centers  for  the  shelves. 

Let  x  =  space  of  the  lower  shelf. 

then  x  —  2=  space  of  the  next  shelf. 

and  x  —  4  =  space  of  the  upper  shelf. 

Adding,   x+  (x  —  2)  +  (x  —  4)  =  total   space. 

Adding,  x+x-2+x-4=  =  32. 

Or,  3x-6  =  32. 

Or,  3x  =  3S. 

Or,  #  =  12§  space  of  lower  shelf. 

Or,  x  —  2  =  10 J  space  of  middle  shelf. 

Or,  x  —  4  =  8f  space  of  upper  shelf. 

(5)  It  is  evident,  therefore,  that  the  letter  stands 
for  a  general  number  while  the  Arabic  symbol  stands 
for  the  particular  number.  We  may  speak  of  (a) 
apples  as  well  as  (5)  apples,  providing  we  understand 
that  the  (a)  is  undetermined. 

(6)  If  we  are  to  use  the  letters  to  any  great  ex- 
tent in  number  operations,  it  is  necessary  for  us  to 
review  the  arithmetical  operations  and  readjust 
our  rules  and  laws  to  accommodate  the  added  sym- 
bols. Taking  these  operations  in  order,  those  of 
addition,    subtraction,    multiplication,    and    division 


LETTER  SYMBOLS  13 

will  be  discussed  first.  The  discussion  will  be  very 
brief,  however,  owing  to  the  fact  that  the  operations 
are  to  be  considered  under  separate  topics. 

Problems 

1  What  are  the  Arabic  symbols? 

2  What  do  they  stand  for? 

3  Does  the  Arabic  symbol  stand  for  a  particular 
or  a  general  number? 

4  When  letters  are  used  as  symbols  do  they 
stand  for  particular  or  general  numbers? 

5  What  does  the  expression  generalized  arith- 
metic mean? 

6  Name  other  points  of  enlargement  and  general- 
ization. 

7  Can  you  state  an  advantage  in  having  a  general 
number  symbol? 

8  What  are  the  definite  number  symbols? 

9  Should  the  algebra  be  a  continuation  of  the 
arithmetic  ? 

10  Do  you  understand  the  distinction  between 
a  number  and  the  symbol  standing  for  it?  Discuss 
this  point. 

11  Suppose  2x  horses  cost  $10.  What  would  x 
horses  cost? 

12  If  (a)  boys  are  to  have  12  apples  divided 
among  them,  what  is  the  share  of  each  boy? 


Lesson  2 

Definitions  and  Rules 

To  facilitate  discussion,  the  following  definitions 
should  be  memorized. 

1  Multiplication  is  expressed  by  using  the  dot 
(.),  and,  in  the  case  of  the  presence  of  literal  factors, 
by  writing  the  factors  together. 

Ex.  3a- 2  =  da  Ex.  aXb  =  ab  =  ab 

Ex.  3a-b  =  3ab  Ex.  x-y-z=xyz 

2  Any  number  or  combination  of  numbers  by 
signs  of  operation  is  called  an  algebraic  expression. 

Ex.  2a-\-b-3 
a  An  expression  of  one  term  is  called  a  monomial. 
b  An  expression  of  two  terms  is  called  a  binomial. 
c  An  expression  of  three  terms  is  called  a  trinomial. 
d  An  expression  of  more  than  two  terms  is  called 
a   polynomial. 

3  An  algebraic  expression  not  separated  by  the 
plus  or  minus  sign  is  called  a  term. 

Ex.  4ab 

2a-\-b  —  3  is  an  expression  made  up  of  three  terms. 

4  Any  one  of  a  group  of  numbers  is  called  a 
j actor  of  their  product. 

Ex.  6ax  has  the  factors  2,  3,  a,  x. 

5  Any  one  of  a  group  of  equal  factors  is  called  a 
root  of  their  product. 

(14) 


LETTER  SYMBOLS  15 

One  of  two  equal  factors  is  called  the  square  root 
of  their  product. 

One  of  three  equal  factors  is  called  the  cube  root 
of  their  product. 

One  of  four  equal  factors  is  called  the  fourth  root 
of  their  product,  etc. 

6  A  small  number  written  at  the  upper  right  hand 
of  a  number  is  called  an  exponent,  and  expresses  the 
fact  that  the  number  is  taken  a  certain  number  of 
times  as  a  factor. 

Ex.  a2  The  a  expresses  the  fact  that  (a)  is  used 
twice  as  a  factor. 

7  Any  factor  or  product  of  two  or  more  factors 
is  called  the  coefficient  of  the  remaining  factors. 

Ex.  3a     (3)  is  coefficient  of  (a) 

(a)  is  coefficient  of  (3) 

(  Note)  We  usually  refer  only  to  the  Arabic  symbol 
as  the  coefficient. 

Ex.  5x2y  5  is  coefficient  of  (x2y) 
5x 2  is  coefficient  of  (y) 
5y  is  coefficient  of  (x2) 
x  2y  is  coefficient  of  (5) 
x 2  is  coefficient  of  (5y) 
y  is  coefficient  of  (5x2) 
(Note)  Ordinarily  only  (5)  is  referred  to  as  the 
coefficient. 

8  The  product  obtained  by  using  a  number  two 
or  more  times  as  a  factor  is  called  a  power. 

Ex.  2-2-2  =  8         (8)  is  the  third  power  of  (2) 

Ex.  a  a  a  a  =  a4       (a 4)  is  the  fourth  power  of  (a) 


16  BRIEF  COURSE  IN  ALGEBRA 

9  Signs  of  aggregation  are  those  which  are  used 
to  signify  that  certain  expressions  are  to  be  considered 
as  unified  wholes.  They  are  the  parentheses  ((  )), 
the  brackets  ([  ]),  the  braces  (\    [),  and  the  vinculum 

Example     2x  +  (3y -f- z)  =  2 

3y-\-z  is  considered  as  a  unified  whole. 
Example     6a  -\-(2b  —  c)  —  3c  =  S 

2b — c  is  considered  as  a  whole. 
Example     (x  —  y)—  4z 

x— y  is  considered  as  a  whole. 
Example     (6m-\-n  —  p)-\-3d—4:a 

6m-\-n  —  p  is  thought  of  as  a  whole. 
Example     (6a  -\-b)—c  =  6 

6a-\-b  is  here  considered  as  a  whole. 

10  If  the  sign  of  aggregation  is  preceded  by  a  (+■) 
sign,  the  quantity  within  is  added;  if  preceded  by  a 
(  — )  sign  the  quantity  is  subtracted. 

11  In  case  the  expression  enclosed  by  the  sign 
of  aggregation  is  to  be  multiplied  by  a  number,  each 
term  of  the  expression  within  is  multiplied  by  the 
number. 

Example  2  (3x + 4)  =  6x + 8 
Example  a(x—y)=ax—ay 
Example     2x  (c  —  d)  =  2xc  —  2xd 

12  The  signs  +,  — ,  X,  ••»,  — ,  are  used  as  they 
have  been  used  in  arithmetic. 

13  It  often  happens  that  letter  symbols  are  given 
particular  values  in  an  expression  to  find  the  value 
of  the  expression. 


LETTER  SYMBOLS  17 

Example — If  a  =  2,   6  =  3,   c  =  4 
To  find  the  value  of  3a-\-b  —  c+4:ab-\-2b 
Substituting  values, 
3(2) +  (3) -(4) +4(2)  (3)  +  2(3) 
Or,   6+3-4+24+6 
Or,  35 

Problems 

1  What  are  the  signs  of  operation? 

2  What  are  the  signs  of  aggregation? 

3  Define   exponent,    power,    root. 

4  Define  coefficient,  literal  coefficient,  numerical 
coefficient. 

5  Define    monomial,    binomial,    trinomial,    poly- 
nomial. 

6  What  new  symbol  for  multiplication  has  been 
introduced  ? 

7  What  is  the  sign  of  equality. 

8  If  a  sign  of  aggregation  includes  certain  terms 
of  an  expression,  what  does  this  signify? 

9  How  is  a  root  indicated? 

10  Do  you  find  that  words  have  the  same  meaning 
as  in  arithmetic? 

11  Discuss  fully  the  relationship  existing  between 
arithmetic  and  algebra. 


Lesson  3 

Discussion  of  the  letter  symbols  with  regard  to  the 
four  fundamental  operations. 

Addition: 

(1)  If  4  is  added  to  3  it  is  understood  that  four 
units  are  to  be  grouped  with  three  of  the  same  kind 
whenever  application  is  made  to  a  problem.  It  is 
a  matter  of  common  agreement  that  the  result  is  a 
group  made  up  of  7  units.  All  Arabic  symbols  are 
names  of  certain  groups. 

(2)  In  abstract  numbers  the  symbols  are  con- 
sidered apart  from  particular  objects,  and  the  state- 
ment is  made  that  4+3  =  7,  a  law  which  holds  for  all 
cases  regardless  of  the  objects  considered. 

(3)  In  adding  letter  symbols  there  is  no  possi- 
bility of  saying  that  (a)  units  added  to  (b)  units 
equals  a  group  named  by  an  Arabic  symbol  because 
of  the  fact  that  (a)  and  (b)  stand  for  undetermined 
numbers. 

(4)  To  add  (a)  and  (b)  then,  it  is  only  possible  to 
indicate  the  sum  a+0,  but  in  case  (2a)  units  are  to 
be  added  to  (3a)  units,  it  is  possible  to  write  the  sum 
as,  2a+3a  =  5a.  This  last  is  true,  owing  to  the  fact 
that,  even  though  (a)  stands  for  an  undetermined 
number,  twice  this  undetermined  number  combined 
with  three  times  this  undetermined  number  equals 
five   times  the  undetermined  number.     In  the  ex- 

(18) 


LETTER  SYMBOLS  19 

pressions  (2a)  and  (3a),  (a)  is  considered  the  common 
factor. 

(5)  We  enlarge  the  rule  for  addition,  therefore, 
to  cover  addition  of  numbers  having  letter  factors, 
(literal  factors).  We  say  that  numbers  having  like 
literal  factors  may  be  added  by  adding  the  coefficients 
of  the  literal  factors  and  writing  the  sum  as  the  co- 
efficient of  the  common  factor. 

Example     5b+4:b  =  9b 

(Note)  (b)  is  common,  so  the  coefficients  (5) 
and  (4)  are  added,  giving  the  sum  (9).  This  number 
is  written  as  the  coefficient  of  the  common  factor  (b) 

Example    6a + 4a  =  1  Oa 

Example     lab  -f  4a&  =  6ab 

(Note)  Here  the  two  factors  (a)  and  (b)  are 
common  and  may  be  considered  as  forming  the 
product  (a)  times  (b)  or  (ab) ,  which  product  is  common 
to  both  numbers.  Therefore,  twice  the  product 
added  to  four  times  the  product  equals  six  times  the 
product. 

Example     2a  ■  -f  4a 2  =  6a  > 

(Note)  In  this  problem  the  exponent  (2)  in- 
dicates that  the  number  (a)  is  taken  twice  as  a  factor, 
so  that  2(aa)-f-  4 (a -a)  =6 (a  a)  That  is,  two  times 
(a -a)  plus  four  times  (a  a)  equals  six  times  (a. a), 
or  6a2, — the  2  indicating  that  (a)  is  taken  twice  as 
a  factor. 

(6)  The  addition  of  expressions  involving  two 
or  more  terms  is  accomplished  by  arranging  the  terms 
so  that  the  terms  having  like  factors  may  be  added 
together. 


20  BRIEF  COURSE  IN  ALGEBRA 

Example — Add    3a  4-46  4-  6c 
and    4a  4- 2&  4-  8c 

Sum  is  7a +66  4- 14c 

( Note)  Here  it  will  be  noticed  that  the  numeri- 
cal coefficients  of  the  like  factors  are  added  and  the 
sum  written  as  the  coefficient  oi  the  like  factors. 

Example     Add    6xy+4w-{-&z 2  and 
4w-\-3z2-\-4xy 

This  problem  must  be  rearranged  so  that  terms 
with  like  factors  are  in  columns. 


6xy-\-4\ 
4x^4-4^ 

w+Sz 2 
w-\-3z2 

Adding, 

lG*y+& 

<jv+\1z2 

Problems 

Add      6x 
4x 

4b 
5b 

8cb 

+  7cb 

Add      3a 
4a 

6c 
10c 

4-4x 
+  8* 

Add      3xy 

-\-4xy 

Scd 
4cd 

3mn 
-\-2mn 

Add    216a 
436a 

649b 
36b 

\6abc 
-\-4abc 

Add  +47* 
+85* 

16y 

+  Uy 

16y2 

lly* 

6y2 

Add      liy 

2\y 

6§* 
4|* 

LETTER  SYMBOLS  21 

5  Add      6y,  4x,  83/,  3x,  and  4y 

mn,  4ww,  6x,  5;y,  3x,  and  2y 
3a,  46,  6c,  14a,  10c,  and  18a 
4a,  +6a,  46,  136,  and  +36 
6 J,  +3d,  4d,  3c,  4c,  and  +  8c 

6  Add     3a+46         2x+3y        4a6+7w 

6a +36         3x+2y         6a6+8ra 

15r+2s  lSw+2x        c+lOd 

6r+3s  16w+3x      3c+  6d 

7  Add      14/+  3k+3m 

16l+Uk+6m 


Lesson  4 

Discussion   of   the  four  fundamental   operations. 
(Continued) 

Subtraction : 

(1)  In  arithmetic,  to  subtract  (2)  from  (6),  we 
determine  the  number  which  added  to  (2)  will  give  (6) 

(2)  In  algebra  the  operation  is  the  same  except  that 
we  can  only  subtract  numbers  having  like  literal 
factors.  (6a  —  3a  =  3a).  We  may  define  subtraction 
of  numbers  having  literal  factors  by  stating  that 
numbers  having  like  literal  factors  may  be  sub- 
tracted one  from  the  other  by  subtracting  the  smaller 
coefficient  from  the  larger  and  writing  the  difference 
as  the  coefficient  of  the  common  factor. 

Example     (3x  —  2x  —  x) 
Example     {5xy  —  3xy  —  2xy) 
(Note)     Here  the  common  factor  xy  is  the  pro- 
duct of  the  factors  x  and  y  which  are  common  to  both 
terms. 

Example     From      5a+4&+6c 

Take       2a+3b+5c 

3a-\-  b+  c 

(4)  Arrangement  must  be  made  as  in  the  case  of 
addition  so  that  the  numbers  having  like  factors 
may  be  in  the  same  column. 

Example     From      6c  d + 4m 2 + Sp 
Take       4cd+3p  +2m* 
(22) 


LETTER  SYMBOLS  23 

(Note)  Here  it  is  necessary  to  re-write  so  that 
the  terms  having  like  literal  factors  may  be  in  the 
same  column. 

6cd+4tm>+Sp 

4:cd+2m7+3p 

Subtracting  2cd + 2m 2 + 5p 

(5)  It  will  be  noted  that  the  conditions  for  sub- 
traction are  the  same  as  for  addition. 

(6)  Terms  having  the  same  literal  factors  with 
the  same  exponents  are  called  "like  terms". 

(7)  Only  like  terms  may  be  added  or  subtracted. 

Problems 

1  From   6  10         16         18         160 
take     3  5  9         12  40 

2  From  4a         16a6     6a2b     9xsy    Sxy 
take     2a  3ab     4a2b  10x*y     Ixy 

3  Subtract  4a+6cd+8d>   from  8a+  10cd+4d2 

4  Subtract  3x+4=y'-\-6zw  from  6x-{-8y'-\-lQzw 

5  Subtract  2m+n+3rs  from  12ra-f-4w-|-8rs 

6  Subtract  16kl+23xiy  from  26kl+40x'y 

7  Subtract    10a>+3b'+6cd+9et    from    15a  '  + 
Sb*  +  llcd-f-14e* 


Lesson  5 

Discussion  of  the  four  fundamental  operations 
(continued) 

Multiplication : 

(1)  To  multiply  numbers  together  having  literal 
factors,  it  is  only  necessary  to  multiply  the  numerical 
coefficients  together  and  indicate  the  multiplication 
of  the  literal  factors  by  writing  them  together. 

Example     2a  ■  4b  —  Sab 

Example     5cd-6cx  =  30c7dx 

( Note)  The  multiplication  of  these  two  numbers 
involves  two  (c)  factors.  Write  30  as  the  coeffi- 
cient of  the  product  of  the  literal  factors  30cdcx. 
There  being  two  (c)  factors,  it  is  customary  to  write 
them  as  (c2),  showing  that  (c)  is  taken  twice  as  a 
factor,  or  30c2dx 

Example     lab  ■  5a 3  b  >  =  20a 4  b 3 

(Note)  Here  5-4  equals  20  for  the  numerical 
coefficient,  and  writing  this  number  as  the  co- 
efficient of  the  indicated  product  of  the  literal  factors 
we  have  20aba3b7 

Here  it  will  be  noted  that  (a)  is  taken  a  total  of  4 
times  as  a  factor,  and  {b)  a  total  of  three  times  as  a 
factor,  so  the  product  is  written  20a  Ab 3 

(Note)  As  stated  before,  the  exponent  indicates 
the  number  of  times  a  number  is  used  as  a  factor. 

(24) 


LETTER  SYMBOLS  25 

(2)  It  will  also  be  noted  that  to  multiply  two 
like  literal  factors  together  it  is  necessary  to  add 
the  exponents. 

Example:  x  *  •  x 2  =  x 6 

(3)  In  case  an  expression  of  two  or  more  terms  is 
to  be  multiplied  by  a  single  term,  the  process  is  as 
follows : 

Example:   Multiply    3x2y-{-4:xy3+3w 
by  Sxy 


9x3y2  +  12x2y*+  9xyw 


(Note)     In  this  case  each  term  of  the  larger 
expression  is  multiplied  by  3xy 

Problems 
Perform  the  following  operations  in  multiplication. 

1  Multiply  4a  by  2 

2  Multiply  16b  by  3 

3  Multiply  14c  by  +12 

4  Multiply  +9ab  by  7 

5  Multiply  +lla2cby  +8 

6  Multiply  4:X-\-3y+z  by  4x 

7  Multiply  3w-\-6z7  +4y  by  5w+2z 

8  Multiply  16a2+4b+c+2d  by  lab 

9  Multiply  +21w+3w3-H£  by  ll  +  2w 


Lesson  6 

Discussion  oj  the  four  fundamental  operations 
(continued) 

Division : 

1  Division  is  denned  as  the  process  of  finding  one 
of  two  factors  when  their  product  and  one  factor  are 
given. 

Example:  Given  the  problem  Sa2b-v-lab 

We  may  determine  the  quotient  by  establishing 
the  number  which  when  multiplied  by  lab  will  give 
Sa7b.     This  number  is  evidently  the  product  of  those 
factors  of  8a  *b  not  contained  in  lab. 
8a*b  =  1.1.2.a.a.b 
lab   =l.a.b 
2,2 . a  —  factors  not  contained  in  l.a.b 
Therefore  4a  is  the  quotient. 

Example     10asb2  +  5ab 
Mentally  it  is  possible  to  determine  the  quotient. 
10-^5  =  2 
a3  -r-a  =  a* 
b*  +  b  =  b 
therefore    1 0a  *  b  ■  4-  Sab  —  la  '*  b 
(Note)     The   fundamental    operations    are    dis- 
cussed fully  in  the  next  section. 

2  To  divide  a  polynomial  by  a  monomial,  divide 
each  term  of  the  polynomial  by  the  monomial. 

(26) 


LETTER  SYMBOLS  27 

Example — (6xy 2  +4*  2y+%x  sy  *)  •*•  2xy 
2xy\6xy2+Ax7y+%x3y* 
3y  +2x    +4x2yz 

Problems 

Perform  the  indicated  division. 
8jc-j-2x,     16y+8y 
14a4-+7a,  18xy++2xy 
16a  2b  -r-  2a,    24a6&4-^8a46 
18a4&6-v-+9a3&>,   40x3:y3-hl0x;yJ 

Perform  the  indicated  divisions. 
15xy,  12a2b,  12a2bc 

3x  6ab  +4tab 

+2Sc*d,  -r-10m4w,         +  180r's* 

ScH  +2m'n  +90rs 


THE  NEGATIVE  NUMBER 
Lesson  7 

We  have  discussed  the  enlargement  of  the  number 
system  by  the  use  of  letter  symbols.  We  will  now 
discuss  another  enlargement  of  the  number  system 
not  by  the  addition  of  new  symbols  but  by  the  qualifi- 
cation of  those  we  have. 

It  is  possible  to  qualify  number  symbols  so  that 
the  same  symbols  may  have  two  meanings.  This 
is  accomplished  by  using  the  signs  (+)  and  (  — ) 
as  signs  of  quality  as  well  as  for  signs  of  operation. 

The  sign  (+)  and  (  — )  may  be  used  other  than  as 
signs  oj  operation.  In  such  cases  the  sign  is  con- 
sidered as  a  part  of  the  symbol  and  is  written  with  it. 
Let  us  take,  for  example,  the  reading  of  the  ther- 
mometer. We  are  all  accustomed  to  speak  of  degrees 
above  or  below  zero,  (zero  being  a  starting  point  for 
measurement),  so  find  it  very  convenient  to  indicate 
ten  degrees  above  zero  by  writing  the  symbol  (10) 
with  the  (+)  sign  preceding  it,  as  follows  (+10), 
and  in  like  manner  ten  degrees  below  zero  with  the 
minus  sign,  (  —  10).  Used  in  this  way  the  signs  do 
not  indicate  operations,  but  simply  that  measurement 
is  made  in  one  or  the  other  of  the  two  directions. 
+  10  degrees  means  ten  degrees  above  zero. 
—  10  degrees  means  ten  degrees  below  zero. 

Another  example  would  be  in  measurement  of  dis- 
tance along  a  straight  line  with  symbols  for  measure- 

(28) 


THE  NEGATIVE  NUMBER  29 

ment  in  one  direction  used  with  the  (+)  sign  connected 
and  symbols  for  measurement  in  the  opposite  direc- 
tion with  the  (  — )  sign  connected. 

-3   -2   -1  0  +1   +2  +3 

The  symbols  stand  for  number  groups  as  they  do 
in  arithmetic,  but  by  the  use  of  the  (+)  and  (— ) 
signs  it  is  possible  to  see  at  a  glance  the  particular 
group  referred  to.  If  (+2),  we  know  that  the  meas- 
urement is  in  one  certain  direction  we  may  agree 
upon,  and  if  a  (  —  2)  the  measurement  is  in  the  op- 
posite direction. 

Another  example  is  to  consider  a  man's  assets  as 
(+)  and  his  liabilities  as  (  — ).  $20  to  be  received 
would  be  written  $+20,  while  ten  dollars  to  be  paid 
out  would  be  written  $  —  10.  In  this  case  it  is  evi- 
dent that  the  man  has  ten  dollars  more  receivable 
than  he  has  payable.  The  balance  would  be  $+10. 
The  absolute  value  remains  unchanged  when  these 
signs  are  used  denoting  quality.  In  order  to  avoid 
confusion,  we  will  consider  addition  as  combination. 

In  the  following  examples  please  note  that  this 
idea    of    combination    is    stressed. 

Ex.  Add  3b  and  -4b  Result  -b 

Ex.  Add  5x  and  —3x  Result  +2* 

Ex.  Add  15c  and  2c  Result  17c 

Ex.  Add  -8d  and  +d  Result  -Id 

It  follows  that  to  express  the  combination  of  two 
or  more  numbers  additively,  it  is  only  necessary  to 
connect  the  numbers  by  the  signs  preceding  them. 


30  BRIEF  COURSE  IN  ALGEBRA 

Ex.  To  combine  +8*  and   —  4x 
+Sx-4x  =  4x 

Here  the  combination  of  Sx  and  —  4x  gives  a  balance 
in  favor  of  the  (+)  units  of  four  units.     To  write 

+8jc  —  4x  =  4# 
is  equivalent  to  subtracting  4#  from  8x. 

Ex.  To  combine  —5a  and  2a. 

( Note)     If  no  sign  precedes  the  symbol  it   is 
understood  that  the  sign  is  (+). 
—  5a+2a  =  —  3a 
In  such  a  problem  as 
Sx  —  4#+  7x  —  5x  —  Sx 
the  process  of  combination  calls  for  a  combination 
of  the  (+)  units,  a  combination  of  the  (  — )  units, 
and  finally  a  balancing  of  the  two  sums. 

There  are  (+10*)  units  and  (— 17x)  units,  giving 
a  balance  of  (  —  7x)  units. 

Supposing  +5a  —  6b  is  to  be  added  to  +46  —  3a 
Arranging         5a — 6b 
-3a+46 
Combining        2a  — 2b 

Here  the  balance  between  (5a)  and  (  —  3a)  gives 
(+2a),  and  the  balance  between  (  —  6b)  and  (+46) 
gives  (  —  2b). 

It  should  be  remembered  that  negative  numbers, 
(minus  numbers)  are  meaningful  only  in  relation  to 
positive  numbers,  (plus  numbers).  Just  as  (up) 
has  no  meaning  except  in  relation  to  (down) ,  or  (east) 
except  in  relation  to  (west),  or  (right)  except  in  rela- 
tion to  (left),  so  this  negative  number  has  no  meaning 


THE  NEGATIVE  NUMBER  31 

except  in  relation  to  the  positive  number.  Inasmuch 
as  they  have  these  opposite  meanings,  to  combine 
additively  is  merely  balancing  them. 

( Note)     It  is  assumed  that  only  numbers  having 
like  literal  factors  may  be  so  combined.     All  other 
combinations  are  simply  indicated.     Ex.  To  combine 
additively  (3a)  and  (—46)  we  simply  write 
3a  -46 

Problems 

1  Add       Sa-2b+c-3d+5x 

-6a+3b-6c+5d-8x 

2  Add        17cd-3ax+10cy 

lScd+Aax—  Scy 

3  Add       lxyz+3ab  —  6cd 

—4xyz  —  5ab+&cd 

4  Add       4a  +  36-  6c 

-3a+  46-  4c 
-8a+146-ll<; 
+3a-  2b+  6c 

5  If  a  man's  wealth  is  represented  by  $  —  500 
what  is  he  worth  after  increasing  it  by  $+800? 

6  If  the  temperature  reading  is   —5°  what  is  it 
after  a  rise  of  7°. 

7  A  ship  is  at  latitude  10°.     She  sails  south  640 
miles.     What  is  her  latitude?     (60  miles  =  1°). 

8  What  is  meant  by 

the  temperature,  +3°,  -6°,  +18°,  -20° 
the  latitude,  +3°,  -16°,  +32°,  -6°,  10° 
the  date,  -162,  +1262,  1896,  -460 


Lesson  8 

Negative   Numbers 
Subtraction : 

In  arithmetical  subtraction  we  simply  diminish  a 
group  named  by  a  symbol  by  some  certain  number 
of  units. 

Ex.     8-2  =  6 
Here  the  group  of  8  units  is  diminished  by  2  units, 
giving  us  a  remainder  of  6  units. 

Suppose  (  —  2)  units  is  to  be  taken  from  (+8) 
units.  The  meaning  here  is  that  the  negative  quality 
of  the  (2)  units  is  taken  from  them  ,thus  restoring  them 
to  positive  units.  So  to  subtract  (  —  2)  units  from 
(+8)  units  the  process  is  one  of  removal  of  the  nega- 
tive quality  before  combination  takes  place. 
Thus  8-(-2)  =  10 
Or  7-(-4)  =  ll 

Or  2a-(-3a)=5a 

Suppose  it  is  desired  to  subtract  (+3)  units  from 
(-7)   units. 

(-7)-(+3)  = 
-7-3=-10 

Ex.     Subtract  2x  —  fy+c  from  —3x+6y  —  &c 
Arranging  terms  so  that   those   with  like  literal 
factors  are  in  the  same  columns, 
-3*+  6y-8c 
+2x-  4y+  c 

-5*+10;y-9c 

(32) 


THE  NEGATIVE  NUMBER  33 

Combining,  we  have  (  —  3x)  and  (  —  2x),  giving 
(-5*);  (+6y)  and  (+4y),  giving  C+lOy);  and  (Sc) 
and  (— c),  giving  (  —  9c) 

Thus  it  is  that  the  rule  in  algebra  is  to  change  the 
signs  oj   the  subtrahend  and  proceed  as  in  addition. 

Problems 

1  From     Sc           Sab         16ao       lSbd  -15rs 
take    —46       —  4a&     —  Sac    —  Sbd  —18rs 

2  From     IScd       -3Sab   -43mn     2Srs  -30xy 
take    —  2Scd       —  40a6       60mn  —  14rs  —  43xy 

3  Two  boys  catch  490  fish.  One  boy  catches 
fifty  fewer  than  the  other.  How  many  does  each 
catch? 

4  How  many  years  between  the  dates  +1462 
and  -530? 

5  How  many  years  have  elapsed  since  —622? 

6  Discuss  the  absolute  value  of  a  number. 

7  An  elevator  goes  from  the  fourth  floor  to  the 
basement.  If  distance  above  the  main  floor  is  + 
and  distance  below  is  — ,  how  would  its  distance  from 
the  main  floor  be  represented? 

8  What  problems  can  you  think  of  in  which  the 
signs  +  and  —  might  be  applied  to  the  solution? 


THE  EQUATION 
Lesson  9 

(1)  The  equation  is  not  a  new  thing.  It  has  been 
used  constantly  in  arithmetic,  as  for  example, 

2+4  =  6 
3-2  =  6,  etc. 

(2)  Its  usefulness  has  been  greatly  increased, 
however,  by  the  use  of  the  literal  symbol.  So  effi- 
cient does  the  equation  become  as  a  method  of 
solution  that  an  eminent  mathematician  has  said 
that  the  equation  is  the  vital  thing  in  algebra. 

(3)  The  usefulness  lies  in  the  fact  that  by  carrying 
through  certain  processes  the  unknown  number  may 
be  expressed  in  terms  of  the  known  numbers,  thus 
giving  a  solution. 

Example:   Suppose  a  problem  is  stated  as  fol- 
lows: twice  a  number  added  to  itself  equals  ten. 
Let  x  =   number 

2x  =  twice  the  number 
Then  3*  =  10 

Or*  =  104-3 
Therefore  x  =  3$ 

(4)  The  expressions  upon  the  two  sides  of  the 
equality  sign  are  called  the  members  of  the  equation. 

(5)  The  members  of  an  equation  being  equal, 

it  follows  that 

(34) 


THE   EQUATION  35 

(a)  Equal  numbers  may  be  added  to  or  sub- 
tracted from  both  members  of  an  equation  without 
changing  the  relationship. 

(b)  The  members  of  an  equation  may  be  mul- 
tiplied or  divided  by  the  same  number  without  chang- 
ing the  relationship. 

Example    If  4  =  4 

Then  4+2=4+2 

Or  4-2=4-2 

Example    If  4  =  4 

Then  4.2=4.2  (multiplying  by  2) 

Or  4^-2=4-^-2  (Dividing  by  2) 

Problems 

1  Why  can  the  equation  be  used  to  greater  ad- 
vantage in  Algebra  than  in  Arithmetic  ? 

2  Solve  for  x  in  the  following  equations. 
2x  =  8 

4*  =  8+4 
3x-2  =  8 
4x+3  =  6+5 
3+2x  =  9 

3  Solve  for  the  letter  symbol  in  the  following 
equations. 

3a  =  9 

4a+a  =  6+4 
3a-2  =  8+6 
2b-3b  =  4:-b 
3s+4  =  8 

4  A  man  had  180  apples  in  two  baskets.  In  one 
basket  he  had  twice  as  many  apples  as  in  the  other. 
Find  the  number  of  apples  in  each  basket. 


36  BRIEF  COURSE  IN  ALGEBRA 

( Note)     Let  the  number  of  apples  in  one  basket 
be  x.     In  the  other  were  180— x 

5  Two  boys  had  together  60  fish.     One  had  3 
times  as  many  as  the  other.     How  many  had  each? 

6  The  sum  of  two  numbers  is  105.     One  number 
is  twice  the  other.     Find  the  numbers. 

7  A  number  added  to  six  times  itself  equals  49. 
Find  the  number. 


Lesson  10 

The  simple  equation  (one  unknown  number) 

(6)  If  the  unknown  number  appears  in  the  equa- 
tion with  no  higher  exponent  than  (one) ,  the  equation 
is  called  a  simple  equation. 

Example    2x+4  =  8     ((*)  has  the  exponent  (1)) 

(7)  The  solution  of  a  simple  equation  involving 
one  unknown  number  consists  of  so  applying  the  laws 
mentioned  in  Part  5  as  to  get  an  equation  having  the 
unknown  number  upon  one  side  of  the  equality  sign 
and  known  numbers  upon  the  other  side  of  the 
equality  sign. 

Example    6x+8  =  10 

Subtracting  (8)  from  each  number, 

6*4-8-8  =  10-8 

Or  6x  =  2 

Dividing  both  members  by  (6) 

Or,  x  =  J 

(8)  The  use  of  the  equation  makes  possible  the 
simple  solution  of  many  problems  which  otherwise 
might  offer  difficulty. 

Example — 

(9)  Suppose  I  have  three  times  as  many  German 
students  as  English  students,  and  have  a  total  of  140. 
To  determine  the  number  of  each, 

Let  x  =  number  of  English  students 
(37) 


38  BRIEF  COURSE  IN  ALGEBRA 

Then  3x  =  number  of  German  students 
Or  4x  =  total 
Then  4x  =  140 
And  x  =  35 
And  3x=105 

Ans.     35  English  students. 
105    German   students. 

(Note)  This  problem  may  be  worked  without 
using  the  symbol  (x),  but  it  is  readily  appreciated 
that  to  use  the  symbol  (x)  simplifies  the  solution. 

(b)  Three  boys  have  a  sack  of  apples  to  divide. 
The  oldest  boy  is  to  have  three  times  as  many  as  the 
youngest,  while  the  second  oldest  is  to  have  twice  as 
many  as  the  youngest.  How  many  does  each  re- 
ceive if  the  sack  holds  66  apples? 

Let  x  m  number  of  apples  youngest  boy  receives 
Then  2x  =  number  of  apples  second  boy  receives, 
And  3x  =  number  of  apples  third  boy  receives. 
Then   6x  =  66. 

Or  x  =  1 1 ,  number  of  apples  youngest  boy  receives. 
2*  =  22,  number  of  apples  second  boy  receives. 
3#  =  33,  number  of  apples  third  boy  receives. 

(c)  Suppose,  four  girls  sell  620  Red  Cross  stamps. 
The  second  sells  60  more  than  the  first,  the  third  sells 
10  more  than  the  second,  and  the  fourth  sells  30  less 
than  the  first.     To  find  how  many  each  sells. 

Let  x  =  number  first  girl  sells 
Then  x+ 60  =  number  second  girl  sells, 
and  a; +70  =  number  third  girl  sells, 
and  x  —  30  =  number  fourth  girl  sells, 
Or  4x  +100  =  total  number  sold. 


THE  EQUATION  39 

Then   4*4-100  =  620, 
Or  4*  =  520. 

#  =  130,  number  first  girl  sells. 

#+60  =  190,   number  second  girl  sells. 
#+70  =  200,  number  third  girl  sells. 

#  —  30  =  100,  number  fourth  girl  sells. 

(Note)  Any  solution  may  be  proved  by  sub- 
stitution of  the  value  of  the  unknown  symbol  in  the 
original   equation. 

Example — Suppose  2#+4  =  8 
Then  2#  =  4 
and  #  =  2 
Substitution  of  (2)  for  (#)  in  the  equation 
2(2)+4  =  8 
4+4  =  8 
8  =  8 

Problems 

1  Solve  the  following  equations  for  (#). 
3#+2a  =  4a 

13#  =  26a 
26+2#  =  6& 
2x-2c+3a  =  Sc+6a 
3a+b+x  =  2b 

2  Solve  the  following  equations  for  (a) 
3a-{-2c  —  d 

Aa-3b  =  6b 
6a+4c  =  & 
66+4#=-2a 
y-4d+3a  =  2x 

3  If  I  pay  2  times  as  much  for  a  hat  as  a  shirt 
and  five  times  as  much  for  a  suit  of  clothes  as  for  the 


40  BRIEF  COURSE   IN  ALGEBRA 

hat  and  pay  $19.50  for  all,  how  much  do  I  pay  for 
each? 

4  If  a  man  has  a  certain  amount  of  money  and 
adds  (a)  dollars  and  then  has  (c)  dollars,  how  much 
did  he  have  at  first?     (Solve  in  terms  of  (a)  and  (c).) 

5  Two  boys  raked  a  yard.  One  received  40  cents 
less  than  the  other.  Together  they  received  $1.20. 
How  much  did  each  receive? 

6  Two  men  go  into  business,  The  first  puts  in 
(a)  dollars  less  than  the  other.  Together  they  put 
in  (d)  dollars.  How  much  has  the  first  man  put  in 
the  business?     (Solve  in  terms  of  (a)  and  (d).) 

7  A  farmer  raised  twice  as  many  bushels  of  oats 
as  corn  and  three  times  as  many  bushels  of  potatoes 
as  corn.  In  all  he  raised  1800  bushels.  How  many 
bushels  of  corn  did  he  raise? 


Lesson  11 

The  simple  equation  (tw.o  unknown  numbers.) 

(10)  An  equation  containing  two  unknown  num- 
bers is  called  an  indeterminate  equation,  owing  to  the 
fact  than  an  indefinite  number  of  pairs  of  values  will 
satisfy  the  equation. 

Example    2x — y  =  8 
Let  x  =  l,  then  2—  y  =  8,  and  y  —  —  6. 
Let  x  =  2,  then  4  —  7  =  8,  and  y=  —  4. 
Let  x  =  3,  then  6— y  =  8,  and  3/=  —  2. 

(11)  It  follows  that  an  equation  containing  two 
unknown  numbers  cannot  be  solved  definitely. 

(12)  It  is  possible,  however,  to  determine  a  pair 
of  values  providing  a  second  condition  is  given  upon 
which  a  second  equation  may  be  built. 

Example    Suppose  2x-{-y  =  16 

Second  condition  x—y=   2 

Adding  equations  3x       =18 

Dividing  by  (3)  #  =  6 

Substituting  (6)  in  either  equation  (y)  is  found 
to  be  equal  to  (4) 

(13)  It  follows  that  a  solution  may  be  determined 
for  two  equations  involving  the  same  two  unknown 
numbers,  but  not  for  one  equation  considered  singly. 

(14)  Such  a  solution  is  called  the  simultaneous 
solution  of  two  indeterminate  equations. 

(41) 


42  BRIEF  COURSE  IN  ALGEBRA 

(15)     Examples: 

(a)    To  solve  j  ?+»"* 

(  3x— ;y  =  4 

Adding  4x  =  10. 

Dividing  by  4,  *  =  2J. 

Substituting  in  the  first  equation,^  2  J -fy  —  6, 

Or  y«3*. 
,,.       (  3a-26  =  4 
W      j  2a-26  =  6 

Subtracting  the  equations,  a  =  —  2 

Substituting,  3(-2)-2&  =  4,  or  6  =  -5 
,  4c-2d  =  4 


3c+3d  =  3 

In  this  pair  of  equations  it  is  noticed  that  neither 
addition  nor  subtraction  of  the  equations  will  eliminate 
either  unknown  number.  By  multiplying  the  first 
equation  through  by  (3)  and  the  second  through  by 
(2),  they  become, 

12c-6d=12 
6c+6d=  6 
Adding  18c         =18 

Dividing  by    (18),   c  =  l 
Substituting  in  the  second  equation,  J  =  0 

(Note)  Solutions  are  proved  by  substitution 
of  the  values  of  the  unknown  symbols  in  the  original 
equations. 


Example — 

Suppose 

x-]~y  =  6 

3x-y  =  2 

Adding, 

4x  =  8 

Or 

x  =  2 

Substituting. 

2+^  =  6,  ;y  =  4 

THE  EQUATION  43 

To  prove,  substitute  these  values  in  the  original 
equations. 

First  equation,  2  -f  4  =  6,  or  6  =  6. 
Second  equation,  6—4  =  2,  or  2  =  2. 

Problems 

1  In  the  equations  used  heretofore,  how  many 
unknown  numbers  have  been  involved?  What  is 
an  indeterminate  equation? 

Determine  five  sets  of  values  for  each  of  the 
following   equations. 

2  3x+2y  =  6 

3  4x-3v  =  8 

4  s+3;y  =  16 

5  -2x  =  10+4^ 

6  14  =  8x-J-2;y 

7  If  one  set  of  values  satisfies  two  equations,  why 
is  this  set  of  values  called  the  simultaneous  solution 
of  the  two  equations? 

Solve  the  following  pairs  of  equations  simul- 
taneously by  the  addition  or  subtraction  method. 

8  3jc+4y  =  8 
2x+5y  =  3 

9  3a+2b  =  6 
4a-&  =  8 

10  4;y-2x  =  6 
3;y-f-*  =  4 

11  2x  =  10-:y 
4*-:y  =  8 


44  BRIEF  COURSE  IN  ALGEBRA 

12  2w-4z  =  6 
4+2w  =  z 

13  6*+4z  =  4 
43+3*  =  5 

Check   results  by  substitution  of    the  values  in 
the    original    equations. 


Lesson  12 

Solve  the  following  problems  using  the  method 
explained  in  Lesson  1 1 . 

1  4x-:y  =  10 
3x+y  =  4 

2  5x+4y  =  8 

x+y  =  2 

3  3*+;y  =  14 

x+2y  =  6 

4  3a+4&  =  6 
4a  =  8+36 

5  3w-\-z  =  3 

z  —  A-Jriv 

6  5a+4c  =  8 
26 -4a  =  6 

Check  results. 

7  Find  two  numbers  whose  sum  is  20  and  whose 
difference  is  4. 

8  Two  pounds  of  butter  and  three  pounds  of 
meat  cost  $.50.  Five  pounds  of  butter  and  four 
pounds  of  meat  cost  $2.30.  Find  the  cost  per  lb, 
of  butter  and  meat. 

9  Twice  the  difference  of  two  numbers  is  4.  Three 
times  their  sum  is  42.     What  are  the  numbers? 

10  In  a  certain  section  of  Wisconsin  the  total 
value  of  the  milk  product  exceeded  the  total  value 

(45) 


46  BRIEF  COURSE  IN  ALGEBRA 

of  the  wheat  crop  by  $300,560.     The  two  were  valued 
at  $422,081.     Find  the  value  of  both. 

11  A  house  and  lot  are  worth  $4,230.  The  house 
is  worth  three  times  as  much  as  the  lot.  What  is 
each  worth? 

12  The  sum  of  two  numbers  is  48.  Their  dif- 
ference is  13.     What  are  the  numbers? 


Lesson  13 

Graphic  representation 

1  If  two  straight  lines  are  drawn  upon  the  plane 
perpendicular  to  each  other  they  may  be  used  as 
lines  of  reference  for  measurement.  For  instance, 
in  the  following  figure  we  may  consider  that  meas- 


V 

X 

0 

X 

1 

1 

y' 

"plaTte'i 

J. 

1 

1 

1 1 1 

urement  to  the   right  of  yy'  is  positive  and   meas- 
urement to  the  left  of  yy'  is  negative.     Also,  we  may 

(47) 


48 


BRIEF  COURSE  IN  ALGEBRA 


1 

TTT 

... 

|       i 

T 

Y 

I       I 

i 

_J 

■ 

1 

!  ! 

I 

' 

5 

1  1 

J- 

5 

U,  i 

,!y=4 

-jU 

Hn 

1 

x' 

c 

X 

ft 

1 

- 

? 

1 

1 

Y 

PI 

Ai 

£ 

n 

1 

, 

BRIEF  COURSE  IN   ALGEBRA  49 

consider  measurement   up  from  xx'<  is  positive  and 
measurement  down  as  negative. 

2  If  the  plane  is  divided  in  squares  and  x  values 
measured  to  the  right  or  left  with  y  values  measured 
up  or  down,  points  may  be  located  having  given 
an  x  and  y  value  for  each. 

Ex.     To  locate  the  points  represented  by 
(1)  *«3      (2)  *«1      (3)  x=-3(4)  x=-2 
y  =  2  y  =  4  y  =  5  y=  —  5 

(1)  To  locate  I  _  ?  1  measure  three  units  to  the 
right  of  the  intersection  on  the  xx'-  line,  then  two 
units  up.  This  locates  the  point.  (2)  To  locate 
the  point  I  ■ .  I  measure  one  square  to  the  right, 
then  four  squares  up.  (3)  To  locate  the  point 
I       _  c    I  measure  three  squares  to  the  left,  then  five 

squares  up.     (4)  To  locate  the  point  i     _  _  -  J  meas- 
ure two  squares  to  the  left,  then  five  squares  down. 

Ex.     Locate  the  following  points. 
(1)     *  =  3  (2)     x  =  6  (3)     *=-2 

y  =  2  y  =  10  y=-4 


50  BRIEF  COURSE  IN  ALGEBRA 

Problems 

Locate  the  following  points  after  drawing  the  lines 
xx'  and  yyr . 

x  =  6  x=-2        x=-3        x  =  3 

<y  =  2  y  =  5  y=~  4        y=— 6 


GRAPHIC  REPRESENTATION 


51 


j 

Y 

,A. 

1 

y=10 

i 

1 

( 

,_ 

l 

[ 

r=i 

x' 

X 

1 

i 

u 

=  -2 

[y=-4' 

Y' 

PI 

At 

E 

n 

-JU 

Lesson  14 

Representation  of  equations 

From  any  given  indeterminate  equation  any  num- 
ber of  pairs  of  values  may  be  found  satisfying  the 
equation. 

Ex.     Given  the  equation  x+2y  =  6 
Ux-l,y-i  If  *--l,  y=i 

If  x  =  2,y  =  2  If**-2/y-4 

If*«3,y«f  If  x=-3,  y  =  j- 

If  *—  4,  y— 1.  etc.  If  x= —4,  ^  =  5.  etc. 

If  these  values  of  X  and  y  are  considered  as  repre- 
senting points,  then  it  is  possible  to  get  a  graphical 
representation  of  the  equation. 

A  line  drawn  through  these  points  represents  the 
equation. 

In  the  above  problem  only  a  few  of  the  infinite 
pairs  of  values  were  located,  but  enough  were  located 
to  lead  one  to  conclude  that  the  line  joining  all  the 
points  would  be  straight. 

Problems 

By  the  above  method  plot  the  following  equations. 

1  *+y  =  2  3  2x+3y  =  S 

2  x— y  =  $  4  x  —  5y  =  6 


(52) 


REPRESENTATION  OF  EQUATIONS 


53 


Y 

V 

S 

<, 

v 

N 

i_ 

x-4,.s: 

y= 

-^ 

i=   -1,  y=. 

!p 

lit, 

=  3 

< 

=  1. 

=2,  y= 

2 

-J, 

y  = 

; 

i 

X 

0 

X 

N**, 

t* 

1 

i 

"S 

Y 

p 

A3 

E 

V 

-L 

Lesson  15 

Representation  of  two  equations 

The  lines  representing  two  equations  will  intersect 
at  some  point  except  when  the  lines  are  parallel. 
This  point  of  intersection  is  the  only  point  that  is 
common  to  the  two  lines;  therefore,  it  represents  the 
simultaneous  solution  of  the  two  equations. 
Ex.  Given  3x+y  =  9 
4x  —  y  =  5 

To  solve  simultaneously  by  graphical  representa- 
tion. 

First,  plot  the  lines  representing  the  equations. 

Second,  determine  the  exact  point  of  their  inter- 
section.    This  is  the  graphical  solution. 

In  the  above  problem  it  is  noticed  that  the  lines 
intersect  very  near  a  point  represented  by  the  values 
x  =  2,  y  =  3.  If  the  equations  are  solved  simultane- 
ously, the  values  of  x  and  y  are  2  and  3.  It  is  evident 
that  there  is  correspondence  between  the  graphical 
and  the  simultaneous  solutions. 

Problems 
Solve  graphically  the  following  pairs  of  equations. 
Then  solve  algebraically  and  compare  results. 
1 


je-|-<y==4 

3       5x+;y  =  16 

x-y  =  2 

x-y=  2 

2x+y  =  3 

4       x+2y=  5 

4x—y  =  3 

3*+4;y  =  ll 

(54) 


REPRESENTATION  OF  TWO  EQUATIONS       55 


-:-:::::::::::-^!:::t:=|::_::::: 

:-:::n::-=::-=z:5E:d-~::::::::- 

==:::=^=±=J=^E|g===:=::::: 

~:::-:=:::::::=-dd-=ffi::-::::- 

:=::::==+:=:="::Iiffi:i=:^: 

-AH  M 

=g""^"^p^ypf:q:==!== 

::::::::::::::#::#:!:::! 

=::::"-~:-::^:::S:::::::r:: 

:~:F::--::::::d::-=:c-:::=:::-:: 

"::::::"-::::-E::::-^::-::::::: 

--:::::-::--:z:3:::::--L:--:-:n;- 

'            f                                                                PLATE  7 

Lesson  16 

A  very  short  method  for  plotting  an  equation  is 
to  give  to  each  unknown  in  turn  the  value  zero. 

Ex.     2x+y  =  Z 

First,  let  x  =  0,  then  y  =  3 

Second,  let  y  =  0,  then  x  =  % 
These  values  represent  points  on  the  axes. 
If  x  =  0,  the  y  measurement  is  upon  the  y  axis. 
If  y  =  0,  the  x  measurement  is  upon  the  x  axis. 
Inasmuch  as  two  points  determine  a  line,  the  graph 
is  determined. 

Problems 

Plot  the  following  equations  by  determining  where 
the  equations  intercept  the  axes. 

1  x+y  =  6  3  x-2y  =  S 

2  2x4-^-5=0  4  3x-2y  =  9 


(56) 


PLOTTING  STATISTICS 


57 


Y 

3 

V 

-t 

\ 

3                                    X 

A                       it 

5             + 

\, '  \~ 

*  £  1  y=s 

\ 

jf                                                                    O      V                                                                  X 

\  fi-*i 

i                                                                        W=tt 

\ 

r 

-U         X     T 

S: 

-£._ 

V 

3                       _|_ 

c  zt        it 

\ 

r 

T*                                                            PLATE  VI 

—„———— Xll_ 

Lesson  17 

Plotting  statistics 

1     Suppose  the  school  attendance  over  a  term  of 
years  is  represented  by  the  following  table. 

1905—496 
1906—500 
1907—480 
1908—510 
1909—525 
1910—550 
1911—580 
1912—590 
1913—615 

The    attendance    curve    may    be    constructed    as 
follows. 


(58) 


|     1              |         -Trr-i "'■" 

1      1 

1       i 

.  j                Zj 

650                                                                                                |                                                     | 

m 

!                           630 

m                                                                 J          i             i   i 

eio                               j                                                1            j        |        j    j 

600                                               /                                                                              j                   !      j      !      i      j      j 

590                                               I                                                                                 [            If     }            |      |      \ 

!     f™                  /            1                         ill!;' 

1            570                                    /                                                                                          1             1            ' 

5«o             /mi     :  i                      !     i     |  j     j 

P               M         I  i  !                       !      i      i  M 

P           /                I                 I     i!       1 

Lo               f           1                                    1  ■     i          • 

ji20                       /                         |      1             1      j                                                                             j 

/             1     M                 I          i 

">  /         1     !     !  I 

•*°~^L-  L  L  L  L  L  L  l  1  '                                i 

s£Ks>5sP5p!§[ii              i 

F  r "  j  r "  r p "    !       i 

1 

!    1         i    |    1    j         ||                         ||                    |||                         IPtAfEVn 

1 1 .1 1 1 1 1 1 1 1         ll,     1  I  1         1 

»                                                                       ».          _ 

j                                                      18                                        y*-.1 

"                         3                                                                                                                                              10 

"""* 

"  J-J__       "       ""       a                                                                                                          Kelt! 

;"T^                   £.'                      —        —               — 

"•  j  dj  S  5  SSJ*S  11  1  i '      jJJi  hil 

Vl    l.;::.."  -XL— 

60  BRIEF  COURSE  IN  ALGEBRA 

2  It  is  desired  to  get  a  mixture  of  milk  and  cream 
which  shall  contain  18%  fat,  from  cream  testing  28% 
and  milk  testing  3% 

In  the  upper  left  hand  corner  of  the  rectangle 
place  the  number  representing  precentage  of  fat  in 
the  cream.  In  the  lower  left  hand  corner  place  the 
number  representing  the  percentage  of  fat  in  the  milk. 
In  the  centre  of  the  rectangle  place  the  number  repre- 
senting the  percentage  to  be  found.  By  cross  sub- 
traction it  is  found  that  15  parts  of  cream  are  to  be 
taken  and  10  parts  of  milk  are  to  be  taken. 

3  A  train  leaves  Chicago  toward  Rockford  run- 
ning 40  miles  per  hour.  A  train  leaves  Rockford  at 
the  same  time  toward  Chicago  running  30 .miles  per 
hour.  At  the  end  of  one  hour,  which  would  mean 
measurement  of  one  unit  up,  the  train  from  Chicago 
has  gone  40  miles,  represented  by  measurement  of 
four  units  to  the  right.  Thus  a  point  is  located 
which  forms  a  straight  line  with  the  original  point. 
Measuring  from  the  opposite  direction  one  unit  up 
and  three  to  the  left,  a  point  is  located  which  forms 
a  straight  line  with  the  original  point.  The  inter- 
section of  these  lines  represents  the  meeting  point 
of  the  two  trains.  The  distance  up  represents  time. 
The  distance  to  the  right  and  left  represents  distance 
from  the  two  cities. 

Note    Chicago  and  Rockford  are  90  miles  apart. 


SECTION  II 

The  Fundamental  Operations 

ADDITION 

Lesson  I 

(i)  The  addition  of  polynomials  depends  directly 
upon  the  addition  of  monomials.  All  that  is  neces- 
sary is  to  arrange  terms  so  that  similar  terms  are 
in  columns. 

Ex.     To  add  3a+2b,  6a+46 

Arranging,    3a + 2  b 
6a+46 
Adding,         9a +66 
Ex.     To  add  Axy  —  6w,  16w  —  3xy 

Arranging,    4xy  —  6w 

—  3xy-\-16w 
Adding,  xy-\~10w 

Ex.    To    add    3ab-2c+4td,    3ab+6c-2d    and 
4c+3d-2ab 

Arranging,    3ab  —  2c+4d 

3ab+6c-2d 

-2ab+4c+3d 

Adding,         4ab-\-&c+5d 

Ex.    To    add    2^-3^+42,    -3x+ 2y-2z,    and 
Sx+2y-3z 

(61) 


62  BRIEF  COURSE  IN  ALGEBRA 

Arranging,    2x  —  3y + 42 

-3x+2y-2z 

5x+2y-3z 

Adding,         4x+  y—  z 

Problems 

1  Add     6x+4^y 

-3x-2;y 

2  Add     3w+4z 

6w  —  2z 

3  Add     2a+4&-  c 

-3a+2b-2c 

4  Add     2a&+5cd 

3ab+8cd 

5  Add      5x;y2  — 2a& 

6x^2  -\-3ab 
2xyz  —  Sab 


6  Add     5x2y+6w 

—  3x2y  —  2w 

7  Add     15m  —  Sn 

2m+3n 
\2m-2n 


Lesson  2 

(2)  Polynomials  having  literal  or  mixed  coefficients 
may  be  added  by  indicating  the  sum  of  the  coeffi- 
cients for  a  new  coefficient. 

Ex.     To  add,  ax+ by,  cx—dy,  and  4x+3y 
Arranging,    ax-\-by 

cx—dy 

4x+3y 
Adding,  (a+c+4:)x+(b-d+3)y 

(Note)     In  this   problem  the  letters  x  and  y 
stand  for  the  variables  or  unknown  numbers. 

(3)  Often  terms  are  to  be  added  involving  binomial 
variables. 

Ex.    To    add,    3(a+x)-\-6(a+x)-2(a+x)  + 
S(a+x) 

Arranging,    3(a+x) 

6(a+x) 

-2(a+x) 

5(a+x) 


and 


Adding,       12(a+x) 

Ex.     To    add,     2x+(a+b), 

4x-2(a+b) 

-3x-6(a+b) 

Arranging,    2x-\-  (a+b) 

4*-2(a+6) 

-3x-6(a+b) 

Adding,         3x-7(a+b) 

(63) 

64  ADDITION 

Ex.  Toadd6w+3w-(*-2;y),3w-2tt+6(*-2:y), 
and  2m-\-n 

Arranging,     6m+3n—   (x  —  2y) 
3m-2n+6{x-2y) 
2m-\-  n 
Adding,       llm+2»+5(#  — 2;y) 
(4)     In  case  expressions  have  terms  connected  and 
held  together  by  signs  of  aggregation,  a  removal  of 
the  signs  of  aggregation  does  not  affect  the  signs  of 
the  terms,  provided  the  signs  of  aggregation  are  pre- 
ceded by  the  plus  (+)  sign. 
Ex.     3x+(±y-3c+d) 
Removing  the  parentheses, 
3x-\-ky  —  3:-\-d 
Ex.     2a-6b+(4x+y-3a) 
Removing    the    parentheses, 
2a  —  6b+4:X-\-y  —  3a 

Problems 

1  Add       ay+bz 

2cy  —  6z 

2  Add     5x-\-aw 

bx-\-cw 
—  2x  —  5w 

3  Add       6a+  4(x-y)-6(x+y) 

10a-  S(x~ y)— 2(*+y) 

2a-10(x-y)+3(x+y) 

4  Combine    6x-4:(a+b)+3c+5x+6(a+b)-4c+ 
8(a+6),  -2x 

5  Combine  15(x+y)  +  20c-10{a-b)+2(x+y)- 
6c+4(x+;y)-2c 


SUBTRACTION 
Lesson  3 

(1)  Algebraic  subtraction  is  the  process  of  deter- 
mining what  must  be  added  to  the  subtrahend  to 
give  the  minuend. 

Ex.     From  lOx  take  3x 
To  find  what  must  be  added  to  3x  to  give  10* 
It  is  evident  from  arithmetical  subtraction  that 
the  answer  is  7x. 

Ex.     From  14a  take  2a 
Answer,    12a 

Ex.     From  6mn  take  3mn 
Answer,  3mn 

(2)  By  reference  to  the  discussion  of  the  negative 
number  in  Section  I,  it  is  evident  that  the  subtraction 
of  a  larger  number  from  a  smaller  gives  a  negative 
number  for  a  difference. 

Ex.     From  7  take  9 
The  answer  then  is  —2 

Ex.     From  2a  take  6a 
If  —4a  is  added  to  6a  the  result  is  2a.     The  answer 
then  is  —4a. 

Ex.    From  3b  take  12& 
If  -9b  be  added  to  126  the  result  is  3b.     -9b  is 
the  answer. 

(65) 


66  BRIEF  COURSE  IN  ALGEBRA 

(3)  The  working  rule  for  subtraction,  when  both 
positive  and  negative  numbers  are  involved,  is  to 
change  the  sign  of  the  subtrahend  and  proceed  as  in 
addition. 

Ex.     From   14cd  take  3cd 

Changing  the  sign  of  the  term  3cd,  it  becomes 
—  3cd. 

Adding  lAcd  and  —  3cd,  we  get  lied 

Therefore  lAcd  —  Zed  —  1  led 

Ex.     From  3xy  take  2xy 

Changing  the  sign  of  the  subtrahend,  2xy  be- 
comes —  2xy 

Adding  3xy  and  —  2xy,  we  get  xy 
Therefore  3xy  —  2xy  =  xy 

Ex,     From  6bc  take   —  2bc 

Changing  the  sign  of  the  subtrahend,  —  2bc  be- 
comes  +2bc. 

Adding  6be  and  -\-2bc}  we  get  Sbc 

Therefore  6bc  -  ( -  2bc)  =  Sbc 
(Note)     From  the  first  general  rule  it  is  evident 
that  it  is  necessary  to  add  -\-Sbc  to  —  2bc  to  give  6bc 

Ex.     From  —  8x;ytake  —  5xy 
Changing   the   sign   of   the   subtrahend,    —  Sxy 
becomes   -\-5xy 

Adding  —  Sxy  and  +Sxy,  we  get  —3xy 
Therefore,   —  Sxy  —  ( —  Sxy)  =  —  3xy 

( Note)  From  the  general  rule  for  subtraction  it  is 
evident  that  it  is  necessary  to  add  —  3xy  to  —  Sxy  to 
get  —Sxy 


SUBTRACTION  67 

Ex.     From  2(c+d)  take   -3(c+d) 

Changing  the  sign  of  the  subtrahend,  —  3(c+d) 
becomes  +3(c+d) 

Adding  2(c+d)  and  +3(c+d)  the  result  is 
5(c+d) 

Therefore,  2(c+d)  - ( - 3(c+d))  =  5(c+d) 

£#.     From  —  &pq  take  6/>g 

Changing  the  sign  of  the  subtrahend,  6pq  be- 
comes —  6pq 

Adding  —  Spq  and  —  6pq  we  get  —14pq 

Ex.     To    simplify    3^+4^  —  2^+^  —  6^ 
Collecting  terms, 

+3*  +4y 

-2*  +  y 

-6jc  

—  5*  +5y        Or  —  5x+5y 

(Note)  In  such  a  problem  the  process  is  one  of 
combination  of  terms  with  consideration  for  the 
signs  preceding  them.  Although  the  negative  sign  is 
involved,  the  problem  is  simplified  by  direct  reference 
to  the  method  of  addition.  (Addition  is  defined  as 
combination  of  terms.)  It  follows  that  the  presence 
of  the  negative  sign  does  not  always  mean  that  the 
process  of  algebraic  subtraction  is  indicated.  In 
all  cases  where  mere  combination  is  to  be  made,  the 
process  is  one  of  algebraic  addition,  even  though  the 
negative  sign  is  involved. 

Problems 

1  Subtract  6a  from  12a 

2  Subtract  2x  from  6x 


68  BRIEF  COURSE   IN  ALGEBRA 

3  Subtract  7  c  from  5c 

4  Subtract  15xy  from  8x3; 

5  Subtract  Amn  from  Smn 

6  Subtract  6a  from  —10a 

7  Subtract  —  5x;y  from  —  Sxy 

8  Subtract  6(a+&)  from  10(a+b) 

9  Subtract  12  (a  —  c)  from  7(a  — c) 

10  Subtract  —15c  from  2c 

11  Combine  3a-f5&-2c+8a-4c-36-18a+ 
Ib-c 

12  Combine    2x-\-y  —  6x-\-4:y  —  3#+4;y  —  7rc+8y 
-3y 


Lesson  4 

Subtraction  of  polynomials 

(4)  The  process  of  subtraction  of  one  polynomial 
from  another  is  performed  by  application  of  the  pre- 
ceding rules  to  each  group  of  similar  terms. 

Ex.     From  3x+2y  take  2x  —  3y 
Arranging,    3x-\-2y 
2x-3y 

Subtraction  of  similar  terms  is  accomplished  by 
changing  the  sign  of  each  sign  of  the  subtrahend,  and 
proceeding  as  in  addition. 

Changing  terms  of  the  subtrahend, 
3x+2y 
-2x+3y 
Adding,  x+5y 

The  process  is  one  of  (1)  arranging  terms  so  that 
similar  terms  are  in  columns,  and  (2)  applying  the 
rule  for  subtraction  to  each  group  of  similar  factors. 

The  terms  of  an  expression  may  be  rearranged  so 
that  similar  terms  are  in  columns  without  altering 
the  value  of  the  expression. 

Ex.     From  3x-\-2y  take  3y-\-6x 
Arranging,    3x+2y 
6x-\-3y 
Subtracting,   —3x  —  y 

(69) 


70  BRIEF  COURSE  IN  ALGEBRA 

Ex.     Perform  the  following  indicated  operation. 

From,         2a+3b 

Take,  a-2b 

Changing  the  signs  of  the  subtrahend, 

2a+3b 

-a+2b 

Adding,  a-\-Sa 

Ex.     Perform  the  following  indicated  operation. 

From,         6x+2y  —  3z 

Take,     -4x+3;y+4s 

Changing  the  signs  of  the  subtrahend, 

6x-\-2y  —  3z 

-\-4x  —  3y  —  4z 

Adding,        10x—  y  —  7z 

Ex.     Perform  the  indicated  operation. 

From,        Sxy —Aw-\-3m2n 

Take,         2xy  —  3w  —  2m2n 

Changing  the  signs  of  the  subtrahend. 

Sxy  — Aw -{-3m  2n 

—  2xy-\-3w-\-2m2n 

Adding,         6xy—  w+5m2n 

Problems 

1  Take   6b-4cd+Sm  from   3b-5cd-10m 

2  Take  lScd2+19m  zn  from  Wed 2 - 1 1  m  zn 

3  Take  6a+3b-3c  from  Sa-Ab+7c 

4  Take  5* —4y-\-6z  from  x  —  2y  —  82 

5  Take    16vw-Sy+6z2    from    Svw-10y-lSzt 

6  From    \Sa-Ud2+c    take    -10a+8d2-3c 


SUBTRACTION  71 

7  From  42xy+3z+Ud  take   14xy-13z-lSd 

8  From  25m5n+4rs2  take  16m5n-Srs2 

9  From  4a&-6cd+14e/  take  7ab+9cd-5ef 
10  From  2a  -  Zed 4  take  -5a+7cdA 


Lesson  5 

(5)  Expressions  involving  literal  coefficients  for  the 
unknown  numbers  may  be  subtracted  by  indicating 
results. 

Ex.     Suppose  x  and  y  are  the  unknown  numbers. 
From,        ax+by 
Take,         cx—dy 

Changing  the  signs  of  the  subtrahend, 
ax+by 
—  cx-\-dy 
Adding,  {a  —  c)x-\-(b-\-d)y 

Ex.     Suppose  m  and  n  are  the  unknown  numbers. 
From,        rm+sn 
Take,      —pm  —  qn 
Changing  the  signs  of  the  subtrahend, 
rm-\-sn 
pm-\-qn 
Adding  (r+p)m+(s-\-q)n 

(6)  Often  the  coefficients  of  the  unknown  numbers 
are  mixed  numbers.  In  such  cases  the  process  of 
subtraction  follows  directly  the  process  of  addition 
after  the  signs  of  the  subtrahend  are  changed. 

Ex.     Suppose  x  and  y  are  the  variables. 
From,        3ax  —  2by 
Take,      —  7ax+3by 

(72) 


SUBTRACTION  73 

Changing  the  signs  of  the  subtrahend, 
3ax  —  2by 
+7ax  —  3by 

Adding ,       1  Oax  —  5by 

Ex.    Suppose  6  and  d  are  the  unknown  numbers. 
From,        3ac+4:bd 
Take,         2xc+3yd 

Changing  the  signs  of  the  subtrahend, 
3ac+4bd 
—  2xc  — 3yd 
Adding,  {3a-2x)c+{4b-3y)d 

(7)  Binomial  and  polynomial  terms  are  treated  as 
monomial  terms  are  treated,  the  coefficients  alone 
being  considered. 

Ex.  From,  2(x-y)+4:(a+b) 
Take,  -3(x-y)+3(a+b) 
Changing  the  signs  of  the  subtrahend, 

2(x-;y)+4(a+&) 

3(*-y)-3(fl+&) 

Adding,         5(x-y)+  (a+b) 

Ex.  From  3a(m-\-n)  —  6b(r—s) 
Take  -2a(m+n)-3b(r-s) 
Changing  the  signs  of  the  subtrahend, 

3a(m-\-n)  —  6b(r  —  s) 

2a(m+n)+3b(r-s) 
Adding,         5a(m+n)—  3b(r— s) 


74  BRIEF  COURSE    IN  ALGEBRA 

Ex.  From,     6ab(p+q+r)-\-3a(m  — y) 
Take,     -2cd(p+q+r)  +  w(m-y) 
Changing  the  signs  of  the  subtrahend, 
6ab(p+q+r)-\-3a(m  — y) 
2cd(pJrq-\-r)  —  w{m—y) 


Adding,    (6ab+2cd)    (p+q+r)  +  (3a-w)   (m-y) 
Problems 

1  From  12a  take  5a 

2  From  lax  take   ax 

3  From  ax  take  bx.  (x  is  the  variable) 

4  From  my  take  ny  (y  is  the  variable) 

5  From  abx  take  cix  {x  is  the  variable) 

6  Subtract  S(a+b)  from  10(a+6) 

7  Subtract  6(x— y)    from    4(x— y) 

8  Subtract  -18(c-d)  from  20(c-d) 

9  From  6  Va6  take  4  Va& 

10  From  8  \/a  take   6  \/a 

11  From  lS(a+b-c)   take  9(a+6-6) 

12  From6(x-2;y+d)   take  S(x-2y+d) 

13  From     3(x+y),        6  Va&,  18(x+y-z) 
Take  -6(x+y),     -7  \/ab,       -24(x+;yr-z) 

14  From     3x2y2z*,     -15  VS    +18  (ro-w) 
Take  -7x2;y223,      -28Vafc,  -25(w-») 


Lesson  6 

Removing  signs  of  aggregation 

(8)  If  signs  of  aggregation,  preceded  by  the  nega- 
tive sign,  are  to  be  removed,  all  the  signs  within  the 
signs  of  aggregation  are  changed. 

Ex.     3ab-2c-(2ab+ 3c) 
To  remove  the  parentheses,  all  signs  of  terms  en- 
closed must  be  changed. 

3ab  —  2c  —  2ab  —  3c 
Collecting  terms,  3ab  —  2ab  —  2c  —  3c 

Or,  ab  —  Sc 

(Note)  It  is  evident  that  the  presence  of  the 
negative  sign  preceding  the  sign  of  aggregation  signifies 
that  the  terms  within  are  to  be  subtracted.  From 
the  working  rule  the  signs  of  the  subtrahend  are 
changed,  and  we  proceed  as  in  addition. 

Ex.     3ab-2c-(6x-2ab+3c) 
Removing  the  parentheses,  and  changing  all  signs 
within, 

3ab  —  2c  —  6x+ 2ab  —  3c 

Collecting    terms,    3ab -f- 2ab  —  2c  —  3c  —  6x 

Or,   Sab  —  5c  —  6x 

(9)  Often  there  are  signs  of  aggregation  within 
signs  of  aggregation.     To  remove  them  the  same  rule 

applies. 

(75) 


76  BRIEF  COURSE  IN  ALGEBRA 

Ex.     3a+[3x+4b-(6a-2x+b)-2a] 
Removing  the  parentheses, 
3a+[3x+4:b-6a+2x-b-2a] 
Removing   the   brackets, 
3a+3x+4b-6a+2x-b-2a 
Collecting  terms, 
3a-6a-2a+3x+2x+4:b-b 
Or,  -5a+5x+3b 

( Note)     It  is  customary  to  remove  the  innermost 
sign  of  aggregation  first. 

Problems 

1  Simplify  16*- (40—  b) 

2  Simplify  8 + 6a  -  (46  -  c) 

3  Simplify  15ef-(6a+3b)  -  (4b -d) 

4  Simplify  13p-  (6r+4) 

5  Simplify  15a+l3b-4d-d+c 

6  Simplify  I6a-(a-b)  +  [c-d]-  \a+d\ 

7  Simplify  a-[b+(c+d)] 

8  Simplify  x+y  —  (a +&  — (w  — w) ) 

9  Simplify  4x+ 1 —4a+3/—(6c+^— 5c)+3y  — 
(4a+c) } 

Enclose  the   following   expressions   with  signs   of 
aggregation  preceded  by  the  negative  sign. 

10  6a+4b-c 

11  8m+16»  — y  —  3r 

Enclose  the  last  three  terms  in  parentheses  pre- 
ceded by  the  negative  sign. 

12  te-3c+5x-3y-±n 

13  18r2s+5mn-16a-4c 

14  2a-4b-16c+lSd+m 


MULTIPLICATION 

Lesson  7 
General  definitions. 

(1)  Multiplication  is  the  process  of  performing  upon 
a  number  (called  the  multiplicand)  the  same  operation 
that  was  performed  upon  unity  to  produce  a  second 
number  (called  the  multiplier) .  The  result  is  called 
the  product. 

Ex.    3-5  =  5+5+5 
The  number  5  is  taken  as  many  times  as  1  was  taken 
to  get  3. 

(Note)     It    is    evident    that    multiplication   is 
shortened  addition. 

(2)  Algebraic  multiplication  follows  arithmetical 
multiplication  directly.  Owing  to  the  enlargement 
of  the  number  system  by  the  use  of  the  letter  symbols 
and  negative  numbers,  certain  revisements  to  the 
arithmetical  rules  are  necessary. 

a    Letter  symbols  may  be  multiplied  in  any  order. 

Ex.    ab  =  ba 
b    Letter  symbols  may  be  grouped  in  any  way 

Ex.     xyz  =  x  (yz)  =  y  (xz)  =  z  (xy) 
c    Polynomial  expressions  may  be  multiplied  by 
monomials  by  multiplying  each  term  of  the  poly- 
nomial by  the  monomial. 

Ex.     (x-\-y—z)>a  =  ax+ay—az 
(77) 


78  BRIEF  COURSE  IN  ALGEBRA 

d     Either  a  negative  multiplicand  or  a  negative 
multiplier  will  give  a  negative  product. 
Ex.     -\-a-  —  b  =  —  ab 
Ex.     —a-\-b——ab 
e    If  both  multiplicand  and  multiplier  are  either 
positive  or  negative,  the  product  is  positive. 
Ex.     a-b=+ab 
Ex.     —a-—b=-\-ab 

(3)  Exponents  from  definition  indicate  the  number 
of  times  a  number  is  taken  as  a  factor,  so  the  law 
follows  that  to  multiply  like  factors,  exponents  are 
added. 

Ex.     23-24  =  2-2-2-2. 2-2-2 
Or,  23-24  =  27 
Ex.     aAa2  =  aaaa  a  a 
OraAa2  =  aG 

Multiplication  of  monomials. 

(4)  Monomials  are  multiplied  by  multiplying  the 
coefficients  together  for  a  new  coefficient,  and  the 
unknown  factors  together  for  a  new  unknown  factor. 

Ex.     3a2b  =  6ab 

Ex.     3xy  ■  4xz  =\2x  2yz 

Ex.     6c24d  =  24c2d 

Ex.     2an-3am  =  6an+m     (adding    exponents) 

(5)  To  multiply  a  polynomial  by  a  monomial,  mul- 
tiply each  term  of  the  polynomial  by  the  monomial. 

Ex.     3x  ■  (2^+3^-4x2)  =  6x  2y+9xy-  \2x  H 

Ex.     2a(4b-c+2d2)=8ab-2ac+4:ad2 

Ex.     6cz(4d-5cd+3e)  =  24c3d-30c*d+18c*e 


MULTIPLICATION  79 

Ex.     2{an-bn)  =  2an-2bm 

(Note)     The  laws  of  signs  are  given  in  working 


form 

l  as  follows: 

1 

Like  signs  give  a  positive  product. 

2 

Unlike  signs  give  a  negative  product. 

Problems 

1 

Multiply  5*  by  2 

2 

Multiply  6ab  by  3a 

3 

Multiply  10x2y  by  2xy 

4 

Multiply    12a*  by  4ay    (Adding  exponents  by 

indicating  operation) 

5 

Multiply  2x+y  by  3x 

6 

Multiply  2*+4  by  5x 

7 

Multiply  4x2+3*+4  by  7x 

8 

Multiply  6a2 -4a -2  by  6a 

9 

Multiply  -2x+4  by  -3x 

10 

Multiply  16s2+4*-5  by  -2x 

Lesson  8 

Multiplication  of  polynomials  by  polynomials. 

(6)  To  multiply  a  polynomial  by  a  polynomial, 
multiply  each  term  of  the  multiplicand  by  each  term 
of  the  multiplier.  Then  arrange  and  add  the  several 
products. 

Ex.     Multiply  3x+y  by  3x  —  y 
3x+y 
3x—y 

Multiplying  by  3x,       9x 2 + 3xy 
Multiplying  by  —  v,  —3xy—y* 

Adding,  9x2— y2 

Ex.     Multiply  3x2+2x-6  by  2x+4 

3x2+2x-6 

2x  +4 
Multiplying  by  2x,       6x 3 + 4x 2  —  1 2x 
Multiplying  by  4,  +  12x2+8*:-24 

Adding,  6x3+16s2-4x-24 

Ex.     Multiply  Scd  -  4  by  2c + 2 
5cd-4: 
2c  +2 
Multiplying  by  2c,       1  Oc  H  -  8c 

Multiplying  by  2,         +  10cd--8 

Adding,  10c2d-8c-r-10cd-8 

(7)  In  multiplication,  it  is  always  advisable  to  ar- 

(80) 


MULTIPLICATION  ,    81 

range  the  terms  of  multiplicand  and  multiplier  so 
that  the  exponents  of  the  letter  symbols  under  dis- 
cussion are  in  ascending  or  descending  order. 

Ex.  Multiply  3x*+2x2-3x*+2-x  by  3x+ 
2x2-3x* 

Arranging  terms, 

3x*-3xz+2x2-x+2 
-3x3+2x2+3x 
-9x7+9x&-6x*+3xi-6xz 

+  6x6-6x5+4:X4-2x3+4xi 
+9x5-9s4+6*3-3*2+6s 

Adding,-9x'J+15x«-3x&-2x4-2x3-x2+6x 

(Note)  The  method  of  arrangement  makes 
possible  the  writing  of  the  products  in  columns,  thus 
simplifying  the  process. 

Definition — The  degree  of  a  term  is  determined  by 
adding  the  exponents  of  the  literal  factors.  The 
degree  of  an  expression  is  determined  by  the  term  of 
highest  degree  in  the  expression. 

Definition — An  expression  is  homogeneous  when 
the  terms  are  all  of  the  same  degree. 

Problems 

1  Multiply  2a  +1  by  3a 

2  Multiply  6x2+3x+2  by  2x 

3  Multiply  6b + 3c  by  4b 

4  Multiply  15m2n  —  2mn  by  mn 

5  Multiply  6ax2- 3a 2x-f-4  by  8x+2 

6  Multiply  Scd-3d2+5  by  2c-3d+2 


82  BRIEF  COURSE  IN  ALGEBRA 

7  Multiply  lOmn— 4m2+2n  by  15w 

8  Multiply  3x  2ys-\-5xzy2  by  6x*y3 

9  Multiply  25ab 2-6a2b  by  Sdb+3 

10  Multiply  14xy+ 1x  2-3y  by   2xyAx  2+y 2 

1 1  Expand  5x(2x  2+4y+3y 2) 

12  Expand  (3a+b)  (6a2b+10) 

13  Expand  (5x+^+z)  (3x2+4;y-3z) 

14  Expand  (10*3+4x2-4x+2)    (6*»+3x-2) 

15  Expand  (6ab*+4ab2-3ab+6a)  (2ab) 


Lesson  9 

Literal  exponents 

1     The  general  rule  for  multiplication  holds  for 
expressions  having  literal  exponents. 

Ex.     {3a2p+2ap+a)  (2ap+l) 
3a2p+2ap+a 

2ap+l        

6asP+4a2p+2ap  +  1 

+3a2p +2ap+a 

6azp+7a2p+2ap+1       +2ap+a 

It  is  essential  to  arrange  terms  in  ascending  or 
descending  powers  to  get  the  best  results. 


Problems 

4xb 

4xa+1+3y 


1     2xa+3y2-±xb 


2  6am-4bn-3c2m 
3a2+4bm 

3  8**-4x*+3 
3xp-x 


4     \5x2ya-4z' 
3xy—zn 


(83) 


DIVISION 
Lesson  10 

(1)  If  the  product  and  one  of  two  factors  are  given, 
it  is  possible  to  determine  the  other  factor.  This 
process  is  called  division.  The  product  is  the  divi- 
dend, the  given  factor  is  the  divisor,  and  the  factor 
found  is  the  quotient. 

(2)  Division  is  expressed  either  by  using  the  sign 
of  division  (-§•),  or  by  writing  the  dividend  and  divisor 
inthe  form  of  a  fraction,  the  dividend  as  the  numerator 
the  divisor  as  the  denominator. 

Ex.     Divide  3ax  by  2y 
Expressed  as  3ax-$-2y 

3ax 

2y 

(3)  To  divide  terms  having  like  factors,  subtract 
the  exponents. 

Ex.     Divide  12x3  by  3x2 

12x3     12-*-x-* 

= =s  4# 

3x2        3  xx 

(  Note)  The  coefficients  are  divided  to  determine 
a  new  coefficient,  and  the  given  factors  are  divided 
to  determine  a  new  literal  factor. 

Ex.     Divide  \6xzy2  by  2x2y 
16x3y2 

(84) 


DIVISION  85 

(4)     The  law  of  signs  is  the  same  as  for  multiplica- 
tion. 

Like  signs  give  a  positive  result. 
Unlike  signs  give  a  negative  result. 

Ex.     -2a2+-a  =  +2a 
+  22x3^  +  llx2=+2:x: 
16x3y-. —  2x 2  =  —  Sxy 
-Sc*d2++4cd2=-2c2     (d*+d*=l) 

Problems 

1  Divide  15a4  by  5a2 

2  Divide  10* 6  by  2x 

3  Divide  20x  4y 2  by  -  lOxy  * 

4  Divide  12m6  by  4ra3 

5  Divide  -  6a  2b 2  by  -  2ab 


Lesson  11 

Polynomials 

(5)  Polynomials  are  divided  by  monomials  by 
dividing  each  term  of  the  polynomial  by  the  mono- 
mial. 

Ex.     2xA-{-Ax3  —  6x2+4x  is  to  be  divided  by  2x 
2x)2x4+4x3-6x2+4x 
xz+2x2-3x+2 

(6)  In  case  the  exponent  of  the  divisor  is  larger 
than  the  exponent  of  the  dividend,  the  new  exponent 
is  expressed  by  the  negative  number. 

Ex.     Divide  3a 4  -  2a  3+4a 2  by  2a 4 
2a4)3a4-2a3+4a2 
|-a-1+2a-2 
( Note)     The  negative  exponent  will  be  discussed 
at  length  in  a  following  topic. 

(7)  Polynomials  may  be  divided  by  polynomials 
as  follows. 

Divide  6x  6+7*  5+6x  4+4s  3+* 2  by2x  2+x 

)3*4+2*3+22+  1 


6*6 

-f-7*5 

+  6x4  +  4xi  +  x2)2x2  +  x 

Multiplying  (2*2  +  x)  by  Sx*  6x6 

+  3x 

S 

Subtracting 

4x5 

+  6x*  +  4x3  +  x2 

Multiplying  (2s2  +  x)  by  2x3 

4xs 

+  2x* 

Subtracting 

4xA  +  4x3  +  x2 

Multiplying  (2x2  +  x)  by  2x2 

4*4+2*3 

Subtracting 

2x>  +  x2 

Multiylying  (2x2  +  x)  by  1 

2x3  +  x2 

Subtracting 

(86) 

DIVISION  87 

Divide  the  first  term  of  the  dividend  by  the  first 
term  of  the  divisor. 

2x 2  is  contained  3x 4  times  in  6x 6 

3x4  is  the  first  term  of  the  quotient. 

Other  terms  of  the  quotient  are  determined  in  a 
similar  manner. 

Illustrative  examples. 

Divide  2a2+7a+6  by  a+2 

)2a+ 3 
2a2  +  7a+6)a+2 
Multiplying  (a+  2)  by  (2a),  2a2  +  4a 

Subtracting,  3a+6 

Multiplying  (a+  2  by  (3),  3aj-  6         quotient  =  2a+  3 

Divide  3xA  —  x3+2x2  —  4x  by  x  —  1 

3x*—x3  +  2x2—4xjx—l 
Multiplying  (x-1)  by  (3x3),  3^4-3a:3 

Subtracting,  2x3  +  2x2  —4x 

Multiplying  (x  —  1)  by  {2x2 ) ,  2x3  —2x2 

Subtracting,  4x2  —4x 

Multiplying  '.x—  1)  by  (4x),  4x2  —Ax 

quotient  =  3% 3  +  2*2  +  4x 

Problems 

1  Divide  4a6+6a4-4a3+2a2  by  2a2 

2  Divide  36a 6 -  27a 4  by  9a 2 

3  Divide  a  2b +ab 2  -  4a  zb 3  by  a& 

4  Divide  a  2+3a+2  by  a+1 

5  Divide  6w 2  —  7w  —  3  by  2w  —  3 


88  BRIEF  COURSE  IN  ALGEBRA 

6  Divide  2x4-6x3+3x2-5x+2  by  x2-3*+l 

7  Divide  6a2-3lab+35b2  by  2a-7& 

8  Divide  2y8-9y2+lly- 3  by  2;y-3 

9  Divide  a2+62+l  +  2a&+2a+26  by  a+b+\ 
10  Divide  #  2+;y2+z2  —  2xy+2xz  —  2yz  by  x—y+z 


Special  Rules  for  Multiplication 

Lesson  12 

Owing  to  the  fact  that  certain  products  recur  time 
after  time,  it  is  of  advantage  to  memorize  certain 
short   methods   for   multiplication. 
(1)     To  multiply  a  binomial  by  itself. 
Given  the  problem  (a-\-b) 2 
B  y  multiplication    a + b 
a+b 
a2+ab 
+ab+b2 


a2+2ab+b2 
This  product  is  made  up  of  the  squares  of  the  two 
terms  of  the  binomial  plus  twice  the  product  of  the 
two  terms. 

It  is  found  that  to  square  any  binomial  the  product 
is  always  equal  to  the  squares  of  the  two  terms  plus 
twice  the  product  of  the  two  terms. 
Ex.     (a-b)2  =  a2-2ab+b2 
Ex.     (x+y)2  =  x2+2xy+y2 
Ex.     (2a-b)2  =  4a2-4cab+b2 
Problems 


1 

ix-y)2 

7 

(a -3b)2 

2 

(c+d)2 

8 

(a-b2)2 

3 

(m-\-n) 2 

9 

(2x2-4y*)2 

4 

{2x+y) 2 

10 

(a+(a-b))2 

5 

(x-2y)2 

11 

(l  +  2x)2 

6 

(2a+3b) 2 

12 

{x2+3a)2 

(89) 


Lesson  13 

(2)    To  multiply  the  sum  of  two  terms  by  their  dif- 
ference. 

Given  the  problem  (a+b)  (a  —  b) 
By  multiplication    a+b 

a  —  b 

a2-\-ab 
-ab-b2 


a2-b2 


It  is  noticed  that  the  product  is  made  up  of  the 
difference  of  the  squares  of  the  two  terms.  To  mul- 
tiply the  sum  of  any  two  terms  by  their  difference, 
the  product  is  found  always  to  equal  the  difference 
of  the  squares  of  the  two  terms. 

Ex.     (x-\-y)  (x  —  y)=x2  —  y2 
Ex.     (x+3y)  (x-3y)=x2-9y2 

Problems 

1  (a-b)  (a-b)  7  (l  +  2a)  (l-2a) 

2  (x+y)  (x-y)  8  (6+*)  (6-x) 

3  (m+n)  (m-n)  9  (2a2+2)  (2a2-2) 

4  (1+x)  (1-x)  10  (4c-d)(4<;+d) 

5  (2a-b)  (2a+b)  11  (2x2-12)  (2*2+12) 

6  (2x-4;y)  (2x+4y)  12  (1 -3a2)  (1-f  3a2) 


(90) 


Lesson  14 

(3)     To  multiply  two    binomials   together  which 
have  a  common   term. 

Given  the  problem  (x+2)  (x+3) 
By  multiplication,  x+2 

x+3 

x2+2x 
+3x+6 


x2+5x+6 


The  product  is  made  up  of  the  square  of  the  com- 
mon term,  plus  the  common  term  with  a  coefficient 
equal  to  the  sum  of  the  unlike  terms,  plus  the  product 
of  the  unlike  terms. 

Ex.     (o+4)  (a-2)=a2+2a-8 
Ex.     (c  +  7)  (6  +  2)=c2+9a+14 

Problems 

1  (a+2)(a+3)  7  (*+6)  (x-6) 

2  (x-4)(x+2)  8  (2a -3)  (2a +4) 

3  (ro+2)(ro-l)  9  (*-2)(*+10) 

4  (*  +  3)(x  +  5)  10  (3,-6)  (y +  5) 

5  (y-l)iy-S)  11  (a2+2)(a2-5) 

6  (r-2)(f+S)  12  (x+ll)(x-8) 


(91) 


Lesson  15 

(4)  Any  two  binomials  may  be  quickly  multiplied 
together  by,  (1)  multiplying  the  first  terms  of  the 
binomials  together,  (2)  multiplying  the  second  terms 
of  the  binomials  together,  and  (3)  multiplying  the 
terms  together  in  such  a  manner  as  will  give  cross  pro- 
ducts. 

Ex.     (2a+4)(3a-2) 
Multiplying  first  terms  together,  6a8 
Multiplying  second  terms  together,  —8 
The  cross  products  are,  12a  and  —4a 
Therefore  product  equals  6a2+8a  —  8 
(Note)     The  cross  products  may  be  easily  de- 
termined by  the  following  process. 

(2a+4)  (3a- 2) 


Ex.     (6c- 2)  (2c+3)  =  12c2+14c 
1 2c  2  =  product  of  first  terms 
—  6  =  product   of   second   terms 
14c  =  sum  of  cross  products 


Problems 

1 

2 
3 
4 

5 
6 

(a+2)  (a+3) 
(x+y)  (w-z) 
(2a -3)  (4a +4) 
(6-3*)  (4+4y) 
(5-2z)(2+4z) 
(l+3a)(6+4a) 

7 

8 

9 

10 

11 

12 

(6-3:y)  (4-4;y) 
(2-*)  (3-;y) 
(46+2)  (3-26) 
(a-4)(2a-2) 
(6+lQ*)(*-3) 
(a+4)(a-26) 

(92) 


Special  Rules  for  Division 

Lesson  16 

The  special  rules  for  division  are  derived  directly 
from  the  special  rules  for  multiplication,  and  lead 
directly  into  factoring.  Factoring  will  be  considered 
as  a  special  application  of  the  short  rules  for  division. 

(1)  If  the  squaring  of  a  binomial  gives  a  product 
made  up  of,  (1)  the  square  of  the  first  term,  (2)  twice 
the  product  of  the  two  terms,  (3)  the  square  of  the 
second  term,  then  any  trinomial,  whose  terms  are  two 
squares  and  a  third  term  equaling  twice  the  product 
of  the  square  roots  of  the  squared  terms,  is  divisible 
by  a  binomial  made  up  of  the  square  roots  connected 
by  the  sign  of  the  second  term. 

Ex.     If  (a+b)2  =  a2+2ab+b2 

a2+2ab+b2 

Then  ;  ~-     =a+b 

a+b 

Ex.     If  (a-b)2  =  a2-2ab+b2 

a2-2ab+b2 

Then : =a-b 

a  —  b 

(  Note)     The  terms  (a  —  b)  and  (a  —  b)  are  factors 

of  a2-2ab+b2 

(2)  A  trinomial  is  a  perfect  square  if  two  of  the 
terms  are  squares  and  the  third  is  equal  to  twice  the 
square  roots  of  the  squared  terms. 

Ex.     a2  —  2ab-\-b2  is  a  perfect  square, 

For,  a2  and  b2  are  squares, 

And  —  2ab  is  twice  the  product  of  the  square  roots. 

(93) 


94  BRIEF  COURSE  IN  ALGEBRA 

Ex.     4c  2-\-4cd-\-d 2  is  a  perfect  square, 

For  4c 2  and  d 2  are  perfect  squares, 

And  4cd  is  twice  the  product  of  the  square  roots. 

Therefore  4c  2+4cd+d 2  is  divisible  by  2c+d. 

Ex.     a2x2-\-2ax+l  is  a  perfect  square, 

For  a  2x 2  and  1  are  perfect  squares, 

And  lax  is  twice  the  product  of  the  square  roots. 

Therefore  a2*2  -f2ax+l  is  divisible  by  ax+ 1 

Determine  which  of  the  following  are  perfect  squares 


1 

x2+2xy+y2 

7 

4c2+8cd+4<22 

2 

a2+4ab+4b2 

8 

x4+x2y2+y* 

3 

x2-2xy+y2 

9 

9a2-6ab+b2 

4 

x2-\-xy+y2 

10 

l  +  2x-\-x2 

5 

25a2+10ab+b2 

11 

16y2+Sy+l 

6 

y2-2\yz+z2    ' 

12 

36w2-24w+4 

Lesson  17 

If  (a+b)  (a-b)=a2-b2 

a2-b2 

Then  —  =  a-b 

a+b 

a2-b2 

And   —  =a+b 

a  —  b 

(3)  The  difference  of  two  squares  is  always  divisi- 
ble by  either  the  sum  or  the  difference  of  the  square 
roots. 

Ex.     c2—d2  is  divisible  by  either  c+d  or  c—d 

c2-d2 

—c—d 

c+d 

c*-d2      ^. 

—  =  c+d 

c—d 

(Note)     (c+d)  and  (c  —  d)  are  factors  of  c2  — d1 

Ex.     4a  2x 2  —  1 6c 2  is  divisible  by  either  (lax + 4c) 

or  (lax  — be) 

4a2x2-16c2     _ 

=  lax  —  4c 

2ax+4c 

4a2x2-16c2     „       ,  , 

=  lax+Ac 

lax  —  4c 

(Note)     (lax+4c)    and    (lax  —  4c)    are   factors   of 

4a2x2-16c2 

Problems 

1  Divide  (x2  —  y2)  by  (#  —  ;y) 

2  Divide  a 4-64  by  a2+o2 

3  Divide  2 2  —  9  by  x  —  3 

(95) 


%  BRIEF   COURSE   IN  ALGEBRA 

4  Divide  x2-9  by  x+3 

5  Divided4  — 16 by  2x2  —  4 

6  Divide  4a  2+4  by  2a  -  2 

7  Divide  9-x2  by  3+x 

8  Divide  16a2-b2by  4a+& 

9  Divide  a2Z>2-4  by  a&-2 

10  Divide  9x*-c2by3x  3+c 

11  Divide  16-4a2  by  4+2a 

12  Divide  a 2-b2  by  a+6 

13  Divide  1-a2  by  1-a 

14  Divide  1-a4  by  1— a2 


Lesson  18 

It  is  found  by  actual  division  that 

%  S-j-'V  3 

=  x2  —  xy-\-y2 

x+y 

xz  —  yz 

And  =  x2-\-xy+y2 

x—y 

It  is  noticed  that  the  sum  of  two  cubes  may  be 
divided  by  the  sum  of  the  first  powers  of  the  quanti- 
ties, and  that  the  difference  of  two  cubes  may  be  divid- 
ed by  the  difference  of  the  first  powers  of  the  quanti- 
ties. 

The  quotients  are  so  striking  as  to  be  easily  re- 
membered. 

(4)  Dividing  the  sum  of  the  cubes  of  two  quanti- 
ties by  the  sum  of  the  quantities,  we  get  the  sum  of 
the  squares  of  the  quantities  minus  the  product  of 
the  quantities. 

(5)  Dividing  the  difference  of  the  cubes  of  two 
quantities  by  the  difference  of  the  first  powers,  we 
get  the  sum  of  the  squares  of  the  quantities  plus  the 
product  of  the  quantities. 

8a3+63 

Ex. -r=±a2-2ab+b2 

2a+b 

8a3-b3 

— -=4a2+2a&+62 

2a  —  b 

(97) 


98  BRIEF  COURSE  IN  ALGEBRA 

Ex.     ^-^=9x2-3xy+y2 

3x+y 

27x3— V3 

— ■ — —  =  9x2+3xy+y2 
3x  —  y 

w  z-\-z 3 

Ex.     =w2— wz+z2 

w-\-z 

w 3— z Z 

=  w2+wz+z2 

w—z 

The  sum  or  the  difference  of  two  cubes  is,  therefore, 
factorable,  and  with  a  short  drill  the  factors  may  be 
determined  by  inspection. 

Problems 

1  Divide  az  —  bz  by  a  —  b 

2  Divide  xz+yz  by  x-\-y 

3  Divide  8 -xz  by  2- x 

4  Divide  27a3-66by  3a-b2 

5  Divide  x6  — v6  by  x2  —  y2 

6  Divide  64  — xz  by  4  —  # 

7  Divide  1  —  # 3  by  1  —  x 

8  Divide  1+y3  by  1+y 

9  Sivide  125+x3by  5+x 

10  Divide  v3-125  by  v-5 

11  Divide  27x3-8  by  3x- 2 

12  Divide  azbz  — dz  by  ab  —  d 


Factoring 

Lesson  19 

The  direct  application  of  the  short  methods  of 
division  is  in  the  determination  of  the  factors  of  an 
expression.  Before  considering  the  special  cases  of 
factoring  it  is  advisable  to  memorize  the  following 
definitions. 

(1)  A  rational  expression  is  one  containing  no  in- 
dicated root  of  the  letter  considered. 

2jc2-|-^+4  is  rational. 

2x2+y/x-\-4:  is  irrational. 

(  Note)  The  sign  V  is  used  to  indicate  a  root 
to  be  found.  The  particular  root  is  indicated  by  a 
number  written  in  the  \/. 

or  v      second  root  (square  root), 
third  root  (cube  root), 
fourth  root. 

If  no  number  is  written  in  the  \J  the  root  is  un- 
derstood to  be  the  second. 

(2)  An  integral  expression  is  one  in  which  the  let- 
ter considered  does  not  appear  in  a  denominator. 

Ex.     3y-f-6  is  integral. 

2 

-4-4^+8  is  not  integral. 

y 

(3)  The  factors  of  an  expression  are  the  expres- 
sions multiplied  together  to  produce  it. 

(99) 


Ex. 

</ 

Ex. 

V 

Ex. 

<f 

100  BRIEF  COURSE  IN  ALGEBRA 

(4)  It  follows  that  the  integral  factors  of  an  ex- 
pression are  the  integral  expressions  multiplied  to- 
gether to  produce  it. 

(5)  Also  the  rational  factors  of  an  expression  are 
the  rational  expressions  multiplied  together  to  pro- 
duce it. 

Problems 

.1  Distinguish  between  the  words  rational  and 
irrational. 

2  What  sign  is  used  to  indicate  that  a  root  is  to 
be  taken? 

3  Define  integral  expression. 

4  Define    factor. 


Lesson  20 

(6)  To  factor  trinomials  which  are  perfect  squares. 

Ex.     a2+2ab-{-b2  (formed  by  squaring  (a+b)) 
Factors  are  (a+b)  (a+b) 
This  follows  from  the  special  rule  for  multiplication. 

(7)  To  factor  the  difference  of  two  squares. 

Ex.     x2  —  y2.    (Product  of    (x+y)    and    (x—y)) 
Factors  are  (x-\-y)  (x—y) 

Problems 
Factor — 

1  x2-2xy-\-y2  4     4x2-4x+l 

2  a2+2ab+b2  5     36a2b2-24abc+4c2 

3  rn2-2mn+n2  6     16x2-32xy+16y2 
7     Factor  the  problems  in  Lesson  16. 

Factor — 

1  (x2-y2)  6  25a2-  16b i 

2  (a4-&4)  7  36a2b2-4: 

3  (x4-;y4)  8  25  -4&2 

4  (4x2-62)  9  9x2-16;y4 

5  (9;y2-l)  10  4w2-9 


(101) 


Lesson  21 

(8)  To  factor  the  sum  of  two  cubes. 

Ex.     xz-\-yz    (Product   of    (x-\-y)    and    (x2  —  xy 

+y2)) 

Factors  are   (x-\-y)    (x2  —  xy+y2) 

(See  special  rule  for  division.) 

(9)  To  factor  the  difference  of  two  cubes. 

Ex.     xz  —  yz    (Product   of    (x  —  y)    and    (x2-\-xy 

+y2)) 

Factors  are  (x—y)  (x2-\-xy-\-y2) 
(See  special  rule  for  division.) 
1     az+bz  7     az-bz 


2 

xz  —  y 3 

8 

1  —  wz 

3 

az  —  cz 

9 

azbz-$ 

4 

8x3+l 

10 

1-c6 

5 

27jc6-73 

11 

a9-64 

6 

1+x6 

12 

a3-l 

(102) 


Lesson  22 

(10)  To  factor  trinomials  made  up  of  two  factors 
having  a  common  term. 

Ex.     x2-x-6.    (Product  of  (x-3)  and  (*+2)) 
Factors  are  (x  —  3)  (x+2) 

The  method  is  to  determine  two  factors  of  the  con- 
stant (in  this  case  —6)  whose  sum  is  the  coefficient 
of  the  first  power  term  (in  this  case  —1). 

Ex.     x2+5x+6.  (Product  of  (x+3)  and  (x+2)) 

Factors  are   (x+3)    (x+2) 
Factors  of  6,  whose  sum  is  5,  are  2  and  3. 

Ex.     a2-5x+6.  (Product  of  (a -2)  and  (a -3)) 

Factors  are  (a  —  2)  (a  —  3) 
Factors  of  6,  whose  sum  is  —5,  are  —3  and  —2. 

Problems 

1  x2+x-6  7  a2x2-6ax+5 

2  x2-\-5x+6  8  4+4x+x2 

3  *2+7x+10  9  a2+12  +  20 

4  a2-3x-10  10  l-2x+x2 

5  y2+15x+26  11  g2+4g_|_4 

6  w2-4«ry+4;y2  12  a2  +  10a  +  16 


(103) 


Lesson  23 

(11)  In  case  the  terms  of  an  expression  have  a 
common  divisor,  this  may  be  removed  and  considered 
as  a  factor. 

Ex.     2a+4a2x+6ay 

2a  is  common  divisor. 
Dividing,  2a(l  +  2ax+3y) 
The  factors  of  the  expression  are  2a  and  1  +2ax+3y 

Ex.     6ayw  —  2za + 4a  2y  3w 
2a  is  common  divisor. 
Dividing,  2a(3yw  —  z-\-2ay  zw) 
The  factors  of  the  expression  are  2a  and  3yw  —  z 
+ 2ay 3w 

Problems 


1 

ax -{-ay 

8 

2x  2y  —  4x  2y 

2 

2b2-b 

9 

a3+4a2-10a 

3 

3c+6c2-9t* 

10 

3xyz+4x  3y 3  —  5y  2s  * 

4 

27a2-9 

11 

5c2+10c 

5 

3xy-\-x2y2 

12 

15y3+6;y2+3:y 

6 

2wz  —  w2 

13 

(a+ft)  2+3(a+b) 

7     4+5a+8  14    4(*-y)  *-2(x-y)  * 


(104) 


Lesson  24 

(12)  If  one  considers  that  the  sign  of  aggregation 
is  used  to  show  that  two  or  more  terms  are  to  be  con- 
sidered as  one  term,  expressions  involving  binomial 
terms  may  be  factored. 

Ex.     (a+b)2+2(a  +  b)  +  l  may  be  divided  into 
the  factors  ((«+&) +  1)   ((a+b)  +  l) 

(Note)     (a+b)  is  considered  as  a  single  term. 
a2+2a+l 

(a+6)2+2(a+&H 
the  same  form. 


>      These   expressions   are  in 


Ex.     (m+2n)2+2(m+2n)   (x-y)  +  (x-y)2 
Factors  are,  [(m+2n)  +  (x  —  y)]   and  [(w+2n)-f- 

(x-y)] 

(Note)     In  this  problem  (m+2n)  and  (x—y)  are 

considered  as  single  terms. 

a2+2ab+b2  [These   ex- 

(m+2n)2+2(m+2n)  (x-y)-\-(x-y)2\  pressions 

are  in  the  same  form. 

Ex.     (a+b)2-(x+y)2 

Factors  are  [(a+b)  +  (x+ y)]and  (a+b)  —  (x+y)] 
a2  —  b2  )  These    expressions    are    in    the 

(a -\-b) 2—  (x+y)  2 )  same  form. 

Ex.     (x+y)  2+5(x+y)+ 6 

Factors  are  [(#+;y)-r-3)]  and  [(#+30+2)] 

a2+5a+6  j  These  expressions  are  in  the 

(x-\-y)  2+5(x+y)  +  6\  same  form. 

(105) 


106  BRIEF  COURSE  IN  ALGEBRA 

Ex.     2(a-2b)+4x(a-2b)-6{a-2b) 
Factors   are    (a  — 2b)    and    (2+4x  — 6)     (Taking 
out  the  common  factor.) 

2a-\-4:xa  —  6a  \  These  expressions 

2(a+26)+4x(a  —  2b)—  6(a  —  2b)  j  are  in    the   same 
form. 

Problems 

1  x2+2x(a+b)  +  (a+b)2 

2  (a-b)2-2(a-b)   (a+b)  +  (a+b)2 

3  (x-y)2-(x+y)2 

4  {2a-b)2-(x-y)2 

5  (46+2)  2-4 

6  (a-b)2+2{a-b)  +  l 

7  (a+6)2+5(a+6)  +  6 

8  (x-3/)2-(x-y)-6 

9  (c+d)2+7(c-d)  +  10 

10  3(a-6)+4(a-6)  +  2(a-6) 

11  4(a-b)2+2(a-b) 

12  6(x+;y)2+9(x+:y)4 


Lesson  25 

(13)     Expressions  may  be  grouped  in  many  cases 
so  that  they  are  factorable. 

Ex.     ax+ay+bx+by 
By  grouping  the  first  two  terms  and  the  last  two 
terms, 

(ax + ay)  +  (bx + by) 

Factoring   each   group,   a(x+y)+b(x+y) 

Considering  this  expression  as  made  up  of  two  terms, 
(x-\-y)  may  be  taken  out  as  a  common  factor. 
(x+y)  (a+b)     The  factors. 

Ex.     2xy + 3axy  —  6xw  —  9axw 
Grouping  the  first  and  third  terms  and  the  second 
and  fourth  terms, 

(2xy  —  6xw)  +  (3axy  —  9axw) 
Factoring  each  group   2x(y  —  3w)+3ax(y  —  3w) 
Considering  this  expression  as  made  up  of  two  terms, 
it  is  factorable. 

(y  —  3w)   (2x+3ax).     The  factors. 

Problems 

1  ax+bx+ay+by 

2  2x-3x-\-2y-3y 

3  4a+66+8a+12& 

(107) 


108  BRIEF  COURSE   IN  ALGEBRA 

4  6x-12  +  18x2-36x 

5  3ax-\-2a+9ax2-\-  6ax 

6  xz  —  x2y-\-xy2  —  y3 

7  3a*-6ya2-a+2y 

8  mn  —  np  —  m-\-p 

9  a2+36-3a-a& 
10  a+2b-4a-Sb 


Lesson  26 

(14)  Often  an  expression  may  be  grouped  to  give 
the  difference  of  a  trinomial  square  term  and  a 
monomial  square  term. 

Ex.     a2+2ab+b2-x2 
Grouping,  (a 2 + 2ab -\-b2)—x2 
Or  (a+b)2-x2 

Considering  (a +b)  as  a  single  term,  this  is  the 
difference  of  two  squares. 

Factoring,   [(a+b)+x)]  [(a+b)-x)]f 
Or  (a+b+x)  (a+b-x) 

Ex.     a2-2ab+b2-\6 

Grouping,    (a 2  -  lab + b 2)  - 1 6 

Or  (a-b)  2-16 
Factoring,  [(a-b) +4)]  [(a-b) -4)] 

Or  (a-6+4)  (a-6-4) 


Problems 

1 

x2+2xy+y2-a2 

2 

a2-2ab+b2-c2 

3 

4x2+4x+l-x4 

4 

c2-(a+6)2 

5 

x2-a2-2a&-62 

6 

a2-2a  +  l-16ra4 

(109) 


110  BRIEF  COURSE  IN  ALGEBRA 

7     a2+2ab+b2-x2-2xy-y2 


8 

m2-2mn+n2-r2-2rs-s2 

9 

4m*n2-c2-4cd-4d2 

10 

(a-b)2-x2-2xy-y2 

11 

\-x2-2x-\ 

12 

x2+2xy+y2-49 

Highest  Common  Factor 

Lowest  Common  Multiple 

Lesson  27 

By  inspection  it  is  possible  to  determine  the  highest 
common  factor  of  two  or  more  expressions. 

Ex.     2ax-\-2xy.        Factors  are  2x (a -\-y) 

a2+2ay+y2.       Factors  are  (a+y)  (a+y) 
The  highest  common  factor  is  (a-\-y) 
(  Note)     To  determine  the  highest  common  factor 

it  is  necessary  first  to  factor  the  expressions  under 

consideration. 

(15)  The  lowest  common  multiple  of  two  or  more 
expressions  is  the  lowest  expression  which  will  contain 
the  expressions.  It  is  determined  first  by  factoring 
the  expressions  and  then  taking  each  factor  the  great- 
est number  of  times  it  appears  in  an  expression. 

Ex.     x2—y2        Factors  are  (x+y)  (x—y) 

x2+2xy+y2        Factors  are  (x+y)  (x-\-y) 
2wx-\-2wy        Factors  are  2w(x+y) 
2w(x+y)  (x—y)  (x+y)  will  contain  the  three  ex- 
pressions. 

(16)  The  working  method  is  to  take  all  the  factors 
of  the  first  expression  and  all  the  factors  of  the  other 
expressions  not  already  selected. 

L.  C.  M.  =  2w(x+y)  (x+y)  (x-y) 
(HI) 


112 


BRIEF  COURSE  IN   ALGEBRA 


Ex. 

2ay-3b) 


(2x+y) 


2ax+4a2y  —  6ab  Factors    are    2a(x-\- 

\x2  —  y2        Factors  are  (2x  —  y)  (2x+y) 
x2+7jc+10         Factors  are   (x+5)    (x+2) 
4x2+4xy+y2         Factors   are    (2x+y) 


x2  —  x  —  6        Factors  are   (x  —  3)    (x+2) 
L.    C.     M.  =  2a(x+2ay-3b)     (2x-y)     (2x+y) * 
(x+5)  (x+2)  (x-3) 

Problems 
Determine  the  H.C.F  and  L.C.M.  of  the  following. 


[-2* 

-j  4x2y 
[  6xy2 

\24ax2 

j  16a2x 

[  4a  3x  3y 

i  a2-b2 

j  a2+2ab+b2 

a2+2ab+b2 

a*-b* 

x2+5x-r-6 

x2+6x+9 

2-ft2 


ax+ay 
x2+2xy+y2 

\  x2y  —  xy2 
8     <j  x2-2x;y+;y2 
[x2-;y2 


a-2 

a2+x-6 
1  a2-4a+4 


a 


8 


I: 


2s2-2*-6 

j  (*-3)2 

[x3-27 


SECTION  III 

FRACTIONS 

Lesson  1 

General  definitions. 

1  An  algebraic  fraction  is  the  indicated  quotient 
of  two  expressions. 

2  The  expressions  are  the  terms  of  the  fraction. 

3  The  numerator  is  the  dividend,  the  denominator 
is  the  divisor. 

Fundamental  laws. 

(1)  The  terms  of  a  fraction  may  be  multiplied  or 
divided  by  the  same  quantity  without  changing  the 
value  of  the  fraction. 

x  ax 

Ex.     If  -  =  w.  then  —  =w 
y  ay 

This  is  evident  when  one  considers  that  the  (a) 
factor  of  the  numerator  contains  the  factor  (a)  of  the 
denominator  once. 


x  a 

Ex.     If  -=w,  then  -=w 

y  y 

a 

(113) 


114  BRIEF  COURSE  IN  ALGEBRA 

y  x       y 

If  -  is  considered  the  divisor,  then  dividing   -  by  - 

a  a        a 

x       a 
is  equivalent  to  multiplying  -  by  - 

a        y 

_    x .  a     xa 

Or  -  ~  =  — 
a  y    ya 

From  the  preceding  law  for  multiplication, 

xa    x 

ya    y 

(2)  By  application  of  the  laws  for  signs  in  division. 
a  +  b  gives  a  (+)  sign. 

—  a-. —  &  gives  a  (+)  sign 

—  a-r-b  gives  a  ( — )  sign 

a  -. —  b  gives  a  (  — )  sign 

a  —a   .  —a    .  a   , 

Or  -  gives  + ,  — -  gives  + ,  — -  gives  - ,  -gives  - . 
b  —b  b  —b 

—  a 
It  follows  that  if  the  quotient  of  — -  gives  — ,  that 

b 

—a    .  t   ,  a    . 

7  gives  -f,  and  that gives  +. 

b  —b 

a     —a         —a  a 

°rr-fr — i"—b 

(3)  From  the  laws  for  removal  of  signs  of  aggre- 
gation preceded  by  the  negative  sign, 

-(x+y)  =  -x-y, 
Or   —  (x— y)  —  —  x-\-y  —  y  —  x, 
Or  —(—x-\-y)=x  —  y., 

It  follows  that  if  a  binomial  factor  appears  in  either 


FRACTIONS  115 

the  numerator  or  the  denominator  of  a  fraction,  chang- 
ing the  signs  of  the  terms  of  the  binomial  changes 
the  sign  of  the  fraction. 

a  —a 

Ex. 


Ex. 


x—y  y—x 

a  —  b  b—a 


a-\-b  a+b 


Problems 

1  Define  fraction,  terms  of  a  fraction,   numera- 
tor, denominator. 

2  Discuss  the  fundamental  laws  governing    the 
fraction. 

3  What  laws  of  signs  hold  for  the  fraction. 


REDUCTION 

Lesson  2 

Reduction  of  fractions  to  equivalent  forms. 

(1)     Applying  the  law  for  division  of  both  terms  of 

a  fraction  by  the  same  quantity. 

6aW         2.3.a.a.b.b.b        2ab2 

3abc  Z.a.b.c  c 

In  the  above  problem,  we  divide  both  terms. by  the 

.      .   .      2a&2 
quantity  3ab,  giving ' 

It  follows  that  to  reduce  a  fraction  to  its  lowest 
terms,  it  is  only  necessary  to  divide  both  terms  of  the 
fraction  by  the  highest  common  factor  of  the  terms. 
42a  2b  zc         6a 


Ex. 


2Sab*c2        4bc 
(Dividing  by  lab  zc) 

_        50(c+d)2         5(c+d) 
'     20(c+d)  2 

(Dividing  by  10(c+d) 

If  the  terms  of  a  fraction  are  made  up  of  polyno- 
mials, (1)  factor  both  terms,  (2)  divide  by  the  highest 
common  factor. 

x4+2x2+\  (x2+l)2  s2+l 

x4-l  (x2-l)(*2+l)     x2-l 

(Dividing  by  (x  2+ 1) 

(H6) 


Ex.    - 


REDUCTION  117 

x3+l         (*+l)  (x2-x+l)     x2-x+l 


x2+2x+l  (*+D2  x+1 

(Dividing  by  (x+1)) 

3x+3b         3(x+b) 3_ 

Ex'    x2+2ab+b2~~  {x+b)2~  x+b 
(Dividing  by  (x+b)) 

(2)     To  reduce  a  fraction  to  an  integral  or  mixed 
expression. 

3ab  „    x2+2xy+y2     (x+y)2 

Ex.—r  =  3         Ex. —  = ; =x+y 

ab  x+y  x+y 

3x  a  —  b 

In  the  above  the  denominator  is  contained  an  in- 
tergal  number  of  times  in  the  numerator. 

6x+y     6x     y  y 

EX-     ^x~  =  2x+Tx  =  3+2x 

a<+l 

Ex.     — — 

a+1 

Dividing  the  numerator  by  the  demoninator. 

(a+1 
a4+l  (az-a2+a 

a4+a3 
-a3+l 

—  n.  3  —  a.  2 


+a2+l 
+a2+a 
-a+1 


118 


BRIEF  COURSE  IN  ALGEBRA 


The  quotient  is  a3  —  a2-\-a  with  a  remainder  of 
—  a+1.     This  remainder  is  placed  over  the  divisor 
and  the  fraction  is  added  to  the  quotient. 
•a+1 


az—a2-\-a- 


a+1 


To  reduce  a  fraction  to  an  integral  or  mixed  ex- 
pression, it  is  only  necessary  to  divide  the  numerator 
by  the  denominator,  and  in  case  of  a  remainder  to 
place  it  over  the  divisor  and  add  the  resulting  fraction 
to  the  quotient  already  found. 

Problems 
Reduce  to  lowest  terms 
tab 
2a 
12x2y 

2  ~zzr 

,    oxy 
18a  2bx 

3ax 
2a 

4   £ 

a2-b2 


8 


a2-2ab+b2 


10 


x2+5x+6 
x2+6x+9 
4x+3 

2x 
6a2+4a+5 

la 
4jc3-|-3jc2+jc+1 

x+1 
5a2+b-c 

5a2+b 


Addition  and  Subtraction  of  Fractions 
Lesson  3 

Addition  and  subtraction  of  fractions. 

( 1 )  To  add  or  subtract  fractions  it  is  necessary  that 
they  have  the  same  denominator. 

x     3x     4x     Sx 
Ex.     -+-  +  -  =  - 

y         y  y  y 

This  is  evident  when  one  considers  that 
x+y+w    x    y    w 
z  z     z     z 

(2)  Fractions  which  do  not  have  the  like  denomi- 
nators may  often  be  written  as  equivalent  fractions 
which  have  the  like  denominators.  They  may  then 
be  added. 

'     3     4     12^12     12 

In  the  above  problem  both  members  of  (|)  were 
multiplied  by  (4),  and  both  members  of  (|)  were 
multiplied  by  (3).  The  resulting  fractions  have  the 
same  denominator  (12),  and  may  be  added. 

To  combine  two  or  more  fractions  so  that  they  may 
have  the  one  denominator  is  called  simplification. 

x     y     w 

Ex.     Simplify  -  +  t 

a     o      c 

(119) 


120  BRIEF  COURSE   IN  ALGEBRA 

x    xbc,     y    yac,        w        wab 

a     abc     b     abc  c  abc 

.   .       xbc + yac  — wab 

Combining 

abc 

a  b 

Ex.     -  +  -7-T- T 
x     x{x-\-y) 

a    a(x+y) 

x     x(x+y) 

a(x+y)+b 


Combining 


x{x+y) 


It  is  best  first  to  determine  the  lowest  common 
denominator  of  the  denominators  of  the  fractions  to 
be  combined.  This  prevents  needless  multiplication 
of  terms. 

^  a  b 

£x 

(*+30 2    x2—y2 
L.C.D.  of  (x+y) 2  and  x2— y2  is  (x+y)2(x  —  y) 
a  a(x—y)  b  b 


(x+y2'     (x+y)2(x-y)      x2-y2         (x+y)(x-y) 

b(x+y) 

(x+y)2(x-y) 

~      -,  .       a(x-y)-b(x+y) 

Combining,  — 

(x+y)2(x-y) 

(  Note)     Time  will  be  saved  if  the  denominators 
of  the  fractions  are  first  factored.     The  lowest  common 
denominator  is  then  more  easily  determined. 
jc+1       x—  1       x+3 


Ex. 


2x+2     4x-4    x2-\ 


FRACTIONS  121 

Lowest  common  denominator. 
2(*+l) 


4(*-l)  Y      4(*+l)  (x-1) 
(*+l)(*-l)J 

Multiplying  each  fraction  by  such  a  number  as  will 
give  equivalent  fractions  having  the  same  denomina- 
tors. 

x+1  2(*+l)(*--l)         2(x2-l) 


2(*+l)  4(*+l)(*-l) 

x-1  (*-!)(*+!) 

4(*-l)     =  4(*+l)  (x-1) 

x+3  4(x+3) 

"  x2-l~  4(x+l)(x-l)'        4(*2-l) 

2(*»-l)  +  (**-l)-4(*+3) 


4(x2- 

-1) 

x2- 

-1 

4(*2- 

-1) 

4(*+3) 

Combining 
Or 


4(x2-l) 
3(x2-l)  -4(x+3) 


4(x2-l) 

(3)  Often  fractions  may  be  combined  more  easily 
if  the  signs  of  the  factors  of  the  denominators  are 
changed. 

la                            lb 
j?x      _l_  

(x+y)  (x-y)         (x+y)  (y-x) 

la  2b 

This  may  be  written    - — ; — — —  . — ■ — — 

(x+y)  (x-y)      (x+y)  (x-y) 

2a -lb 


(x+y)(x-y) 


122  BRIEF  COURSE  IN  ALGEBRA 

Problems 


Add. 

1 

2    3 
re'  2x 
4a    3a    5a 

2 

2      3      4 

3 

6x,                4x 

a2-62    a2-2a&+62 

4 

(a  —  b)          a+b 

a2-2ab+b\     a-b 

5 

3,         5a 

2x    4x2+6x 

5a      —  6a      \a 

6 

2'        3    '      4 

7 

3(a-6)            2a+4& 

2x  — ^       4x2  —  4xy-\-y2 

8 

10,              5 

a— 6         a+& 

Combine 

9 

3               2              4 
ax            ax2            x 

10 

2                   4 

6 

a-b            a2-b2     ^     c 

• 

11 

5x               3x               4* 
2x+l              2              2x- 

-1 

12 

x+1                  x-2 

2 

s2-4x+4            x2-4 

X 

FRACTIONS  123 

2a  4a  6a 

13  aJ-b1*     +     a-b     ~     a*+ab+b2 

2  6  4 

14  -     -     zr    + 


15 


*  3*  2*+4x2 

a  — 6  a+& 


a2+7*-fl0  a2+8s+12 


Multiplication  of  Fractions 
Lesson  4 

(1)  As  the  numerator  stands  for  the  number  of 
equal  parts  taken,  it  is  evident  that  to  multiply  the 
numerator  of  a  fraction  by  any  number  is  equivalent 
to  multiplying  the  fraction  by  that  number. 

1  2 

Ex.    -      2   -   -  =   1 

2  2 

2  6 

Ex.    -   •   3   =  -   =   2 

3  3 

a  a2 

Ex.    -  •  a   =   y 

(2)  As  the  denominator  of  a  fraction  is  the  divisor, 
it  is  also  evident  that  to  multiply  the  denominator 
of  a  fraction  by  any  number  is  equivalent  to  dividing 
the  fraction  by  that  number. 

Ex.     i  is  divided  by  (2)  by  multiplying  the  de- 
nominator (2)  by  (2).     (J)  is  one  half  of  (£). 

Ex.     j  is   divided   by   (2)   by  multiplying  the 
denominator  (4)  by  (2). 

(3)  To  multiply  a  fraction  by  a  fraction,  it  is  only 
necessary  to  multiply  the  numerators  together  for  a 
new  numerator  and  the  denominators  together  for  a 
new  denominator. 

(124) 


FRACTIONS  125 

EX.      f    .    i   m   | 

J  is  to  be  taken  i  times.  This  means  that  the 
fraction  is  to  be  divided  by  2. 

In  the  problem,  f  .  f 
i  is  taken  J  times,  which  means  that  the  fraction  is 
first  to  be  divided  by  (3)  and  then  (2)  parts  taken. 
To  divide  f  by  (3),  multiply  the  denominator  by  (3). 
To  take  (2)  of  these  parts,  multiply  the  fraction  by 
(2)  by  multiplying  the  numerator  by  (2). 

6a2xy       m  18*  Vg2 

Ex.     •   oxv  =  

7a3xz2  J  7azxz2 

(Multiplying  the  numerator  by  3xy) 

6a2b      8a*b       2xab        96a«b*x 

Ex.     • •    =   

4c         5a2        lac  A0c2a3 

(Multiplying  the  numerators  together  for  a  new 
numerator,  and  the  denominators  together  for  a  new 
denominator.) 

x+1      x-1       (x+i)(*-l)         xz-l 
Ex. 


2*-3         6  6(2*- 3)  12s-18 

(Multiplying  the  numerators  together  for  a  new 
numerator,  and  the  denominators  together  for  a  new 
denominator.) 


Problems 


2 

1  Multiply  -  by  3 

x 

4a 

2  Multiply  —  by  5 

ZiX 


126  BRIEF  COURSE  IN  ALGEBRA 


2y 

3  Multiply  ~  by  8 

o 

a-b        1 

4  Multiply  — —  by  - 

5  Multiply^  by  ^ 

a  —  b  a 

6  Multiply  — —  by 


a+b 

7  Multiply  —  by  — 

1—  jc  1 

8  Multiply  ~—  by  — — 

9  Multiply  ~  by  £ 

10  Multiply^  by  ~j 

2x+2 

1 1  Multiply  — - —  by  3x 

««     w  ,  .  ,    3x2+2*-l  3* 

12  Multiply  -       — -     by  — • 

X  —  L  X-\-6 

(a-b)2        a-b 

13  Multiply  L^L  by  — 

(a-b)2  a-b 

2  4s  — 2 

15     Multiply r  by  — - — 


FRACTIONS  127 


16  M^^KZlXx+\°5hy(^=r 

Note.    Factor  and  cancel. 

x*+yz  (a+b)2 


17 
18 


a2+2ab+b2      x2+2xy+y2 

ax-\-bx  a2  —  b2 

a2+2ab+b2  '        x 


Division  of  Fractions 
Lesson  5 

a  x 

If  t  is  to  be  divided  by  -\ 

o  y 

a     x 

j--7-^  =  z     (where  z  is  the  quotient.) 

Multiplying  the  quotient  by  the  divisor  gives  the 

dividend. 

_    a        x 

Or  r  =  z.~. 

o        y 

y 
Multiplying  both  members  by  -. 

x 

ay        x  y 

r--  =  z.-.- 

.     ox        y  x 

r\  a-y 

Or  T  -  =  2 
b  x 

The  conclusion  is  that  to  divide  a  fraction  by  a  frac- 
tion, one  may  invert  the  divisor  and  multiply. 

Ex.     To  divide  — —  by  - — 

56       J  2c 

6a2     dab     6a2  2c        12a  2c 


56       2c       5b     Sab      SOab2 

6x2+3y     2x-y        6x2+3y    3x2+y* 

£Lx.  .  -7- = .  • = 

4*-2        2>x2+y7      4*-2       2x—y 

(6s2+3;v)(3*2+;y2) 
(4x-2)(2x-y) 

(128) 


FRACTIONS  129 


(2  +  **)  +(2x-~\r  2y+3x  '  2xy-6x* 


y 

2y+3x y_  2y+3x 


'2xy-6x2      2xy-6x2 
Problems 


x 
2a 

Tb'6' 

a  —  x^l 
~3~  '.2' 

2+a^_a+b 
a—b  '  a  —  b 

1        a+b 


9+c     9-c 

6  a2-b2     ^_a+b 
a2+2ab  +  b2  '  a  —  b 

Note    Cancel  whenever  possible. 

7  *2+ll*+30  ^s+6, 
x2  +  10  x  +  25  *  x+5 

Q     2x+4     x+2 

3a2  +  6a4  +  18a2  3a2 


130  BRIEF  COURSE   IN  ALGEBRA 

(a-b)3  ,a2-2ab+b2 


10 
11 
12 


4*  12* 

18a3         6a 


a  +  b     a2—b2 

2     ^2a2-4a3 

a—x  '    (a—x)2 


Complex  Fractions 
Lesson  6 

1  A  complex  fraction  is  one  in  which  fractional 
terms  are  involved  in  one  or  both  its  members. 

To  simplify  a  complex  fraction,  simplify  each  mem- 
ber separately. 

Ex.     Simplify 


a 

1+i 


a  a     _       b  +  a  _       b      _  ab 

a     b  +  a  b  b  +  a      b  +  a' 

1  +  b    T 


Ex.     Simplify     b 


•  t  i+(i+p  i+d.|) 


a 


a  a 

b  b 


b    a  +  b    ba  +  b    ab+b2 

1+-    

a       a 

x     c 

Ex.     Simplify  — • 

c     * 

d~2y 
x     c       xd+yc 

yd  yd         xd+yc     2cy  —  dx     xd+yc 

c      x  =2cy—dx=    yd  2yd  yd 

d~2y         2dy 
2dy     _2xd  +  2yc 

2cy—dx     2cy—dx 

(131) 


132  BRIEF  COURSE  IN  ALGEBRA 


y 

2-- 

x 

Ex.     Simplify    - 

y+- 

*    x 

y     2x-y 

x        x        2x—y     xy+1     2x—y     x        2x—y 


1     %y  +  l        xxx     xy+1     xy+1 

y+- 


x 

1 


Ex.     Simplify     1+ i 

1  + 


x 
1 


1    - 


1  1  1         =1^_2X  +  1_ 

„  ,      *        *  +  !+*     2*4-1         '   x  +  l  " 


ac-4-1       #  +  1         x  +  1 
j    #4-1       flp+l 


2*  +  l~2*  +  l' 

Problems 
Simplify 

i         »  3       2a~4&       5 

__         3.     -  _     5. 


x  '  l+o 


•+1  3,    1 


2+3* 
1-* 

4-* 
'1+x 

3      , 

S-x 

2+*  ' 
1 

2 

1 

14- 

2*-J_     4-     _*         6- 

x  x 


Fractional  Equations 
Lesson  7 

Solution  of  equations  involving  fractions. 

To  solve  an  equation  involving  fractions,  (1)  clear 
the  equation  of  fractions,  (2)  solve  by  methods  before 
mentioned. 

L.C.M.  of  3,  2,  6  =  6. 

Writing  all  terms  as  fractions  having  the  common 
denominator, 

4y     15y_3y     48 

T    e     e    o" 

Multiplying  the  equation  by  (6), 

4;y+ lSy  =  3y+48. 

Collecting   terms, 

16^  =  48. 

Or  y  =  3. 

„        „     ,3x+4     6x-2 
22*.     2x+  — - —  «  — — - 
4  8 

L.C.M.  of  4,  8  =  8 

Writing  all  terms  as  fractions  having  the  denomi- 
nator (8), 

16s     2(3*+4)_  (6* -2) 

1T+       8  8       * 

Multiplying  the  equation  by  (8), 
Simplifying,  1 6x + 6x + 8  =  6x  —  2 
Collecting  terms,  16x  =—10 

Or,  *=-f 

(133) 


134  BRIEF  COURSE  IN  ALGEBRA 


Ex.     -L-    4  8 


x  +  l     x-l     x2-l 
L.C.M.  of  (x+l),   (x-1),  and  (x2-l)  =  (x+l) 
(x-l)=x2^-l 

Writing  all  terms  in  equivalent  form  having  the 
common  denominator. 

3(*-l)  _4(*+l)_      8 
x2-l   "   x2-l       x2-l 
Multiplying  the  equation  by  (x2  —  1), 

3(*-l)-4(*+l)  =  8 
Expanding  and  simplifying, 

3x  —  3  —  4x  —  4  =  8 
Collecting  terms,   —  #  =  15 
Or  *=-15 

Ex.     Solve  for  (x)  when  (a,  c,  b)  are  considered 

as  known  values. 

x     2x  _c     x 

a     cb     b     a 

L.C.M.  of  a,  cb,  b  and  a  —  acb 

Writing  all  terms  in  equivalent  form  having  the 

denominator  acb, 

cbx     lax  _ac2     cbx 

acb     acb     acb     acb 

Multiplying  by  acb, 

cbx + lax  =  ac2-\-  cbx 

Collecting  terms, 

cbx + lax  —  cbx  —  ac 2 

Factoring,  x(cb -\-la  —  cb)  =ac2 

ac2  c2 

Dividing  by  (cb  +  la  —  cb),x 


cb  +  2a—cb      2 


FRACTIONS  135 

Problems 


Solve, 


3,4  ,  a-2     a+3 

x+2~x=6  7  -r~  ~ir=6 

2x_5*     3x  =  4  «  *+3     *+2     x+5 

6     T+~3~  3 

2x+3=| 

*             „  „  (a— bV     a—b     x 

-+2X  =  6  10    1-J 2~  =  3 


5* +3     2*-3 
a+1  +  2a-l 


11 


2a 

3a        5 

3* +2 
6 

6*-4 
+      4 

(a-6) 

2     a-6 

3 

2 

x+3 

*  +  l 

3a-2 

~3a+3 

3a +2 

2a 

11        11 

6     2y+y  =  2+3+^y    M 

13  3ax+6a  =  2*+10a 

x     2x        x        Sx 

14  -+— = -  — — 

a      o      a—b     a 

15  Two  men  were  in  business  and  made  a  profit 
of  $3,225  in  one  year.  One  man  received  20%  more 
than  the  other.  What  part  of  the  profit  did  each 
receive  ? 

16  A  parent  offered  a  boy  one  dollar  if  the  average 
of  his  marks  was  80%  or  over.  If  the  boy  gets  90% 
in  mathematics,  85%  in  history,  78%  in  rhetoric, 
and  70%  in  spelling,  what  must  he  get  in  agriculture 
to  have  an  average  of  80%. 


SECTION  IV 
POWERS  AND  ROOTS 

Lesson  1 

General  definitions. 

1  A  power  is  a  product  obtained  by  using  a  number 
two  or  more  times  as  a  factor. 

Ex.     The  second  power  of  2  =  2  •  2  =  4 
The  third  power  of  3  =  3-3-3  =  27 

2  The  laws  of  signs  hold  as  they  do  in  multiplica- 
tion. 

Ex.     +2  -+2  =  +4 

-2--2  =  +4 

3  The  exponent  is  a  small  number  written  at  one 
side  and  above  a  given  number  to  indicate  the  number 
of  times  the  given  number  is  to  be  used  as  a  factor. 

Ex.  3  2,  2  is  the  exponent. 

.  32  =  9 

Ex.  23  =  8.     3  is  the  exponent. 

Ex.  a2  2  is  the  exponent. 

4  The  process  of  multiplication  performed  on  like 
factors  having  equal  or  unequal  exponents  is  per- 
formed by  adding  exponents. 

Ex.  32-34  =  36 
Ex.  a2-az  =  ah 
Ex.     4°-46  =  4°+b 

(136) 


POWERS  AND  ROOTS  137 

5  The  above  law  is  explained  as  follows: 

a2a3  =  a-aaa'a=(a)   taken   five  times    as    a 
factor. 

This  is  expressed  as  a 5 
b*bA  =  b7  =  b-b-b-b'bbb  =  (b)    taken    seven 
times  as  a  factor,  and  is  expressed  as  b 7. 

6  In  case  the  terms  have  more  than  one  factor, 
the  exponents  of  like  factors  are  added. 

Ex.    a3x2a2xb  =  a5x7 
b2y3-by  =  b3yA 
x2y2z4-x3y5z*  =  x6y7z10 

7  Should  there  be   coefficients  to   one  or  both 
terms,  these  become  factors  of  the  product. 

Ex.     3a2b-a3b  =  3a5b2 
Ex.    4a3b2'5ab3  =  20aAb5 

8  The  root  of  a  number  is  one  of  a  certain  number 
of  equal  factors  of  a  number. 

Ex.     2  is  a  root  of  4. 
Ex.     3  is  a  root  of  27 

9  The  square  root  is  one  of  two  equal  factors. 
Ex.     2  is  the  square  root  of  4. 

The  cube  root  is  one  of  three  equal  factors. 

Ex.    4  is  the  cube  root  of  64. 
The  fourth  root  is  one  of  four  equal  factors. 

Ex.     2  is  the  fourth  root  of  16. 

10  The  sign   V  is  called  the  radical  sign,  and 
indicates  that  a  root  is  to  be  found. 


138  BRIEF  COURSE  IN  ALGEBRA 

11     A  number  placed  in  the  V  indicates  the  root. 
Ex .     \/  indicates  square  root . 
Ex.     <\/  indicates  cube  root. 
Ex.     \/  indicates  fourth  root. 
This  number  is  called  the  index  number. 
(Note)     The  square  root  is  usually  expressed  by 
writing  the  sign  with  no  number  in  the  V- 
Ex.     Va',  V^,  Vl6 

Problems 

1  Define  power,  exponent,  radical. 

2  Multiply  3x  2y  by  2y 

3  Multiply  4a  2b2  by  lab 

4  Multiply  lOab*  by  bz 

5  Multiply  dm  2n  by  mn 

6  Multiply  m  3n 2  by  Smn 2 

7  Multiply  6x2y  by  3xy2 

8  Express  the  square  root  of  3a +6 

9  Express  the  cube  root  of  4#  —  2y 

10  Express  the  fourth  root  of  3xy  —  4 

11  Multiply  16x  by  3xy 

12  Multiply  4m2  by  In2 


POWERS 

Lesson  2 

Powers  of  monomials 
Integral  exponents. 

1  If  a  number  having  an  exponent  is  to  be 
raised  to  a  power,  the  exponent  of  the  number  is 
multiplied  by  the  number  expressing  the  power. 

Ex.  (a2)3  =  o6 

Ex.  (a3)4  =  a12 

Ex.  (ax)2  =  a2X 

Ex.  (ax)y  =  axy 

2  The  explanation  of  the  above  law  is  as  follows : 

a2  =  aa 

(a -a)  3  =  aaaaa-a  —  a6 

3  If  the  product  of  two  numbers,  each  of  which 
has  an  exponent,  is  to  be  raised  to  a  power,  the  ex- 
ponent of  each  factor  is  multiplied  by  the  number 
indicating  the  power. 

Ex.     (a263)2  =  a4&6 

Ex.     (xzy4)*  =  x9y12 

Ex.     (4a2fc3)4  =  44a8&4c12 

4  The  above  is  evident  when  one  considers  that 
a  2b 3  is  equivalent  to  aabbb.  To  take  this  value 
twice  as  a  factor  is  to  take  (a)  four  times  as  a  factor 
and  (b)  six  times  as  a  factor. 

5  The  process  of  raising  an  expression  to  any 
power  is  called  involution. 

(139) 


140  BRIEF  COURSE  IN  ALGEBRA 

6  Involution  has  to  do  with  the  raising  of  mono- 
mial, binomial,  and  polynomial  expressions  to  powers. 

7  The  raising  of  any  monomial  to  a  power  is 
accomplished  by  raising  each  factor  of  the  monomial 
to  the  required  power. 

Ex.  a2  raised  to  the  second  power  is  (a2)  2  —  aK 

Ex.  2a  raised  to  the  second  power  is  (2) 2 .  (a) 2  = 
4a2 

Ex.  (3azb)2  =  9a«b2 

Ex.  (2xy2)*  =  Sx3y« 

Ex.  (4a-2xn)2=16a-ix2n 
( Note)     In  case  there  is  no  literal  factor,  the  number 
is  not  usually  factored. 

Ex.  42  =  16 

Ex.  53  =  125 

8  The  same  method  is  followed,  regardless  of  the 
seeming  difficulty  of  the  problem. 

Ex.     (x  2y?) 3  =  x  6y* 

Ex.     (4a°&-3)2  =  16&-6     (Note)     o°=l 

9  To  raise  a  fraction  to  a  power,  raise  both  mem- 
bers of  the  fraction  to  the  power. 

Ex.    (§>-*       E~offtV4|-| 

V3/       9  V3/       3  3     9 

(a\  3     a  3 
b)     =  b* 


Ex. 
Ex. 


4s_4 
9y* 

10     Monomials  having  fractional  factors  may  be 
expanded  by  using  the  above  rule. 


POWERS  141 


Ex.      Aa2fc3Y=ja40 

(2         \»      8 
-a6bc  J    =^18< 


3/- 3 


11     In  case  a  monomial  is  to  be  raised  to  a  power 
expressed  by  a  literal  exponent,  the  same  rules  hold. 
Ex.     (a2)w  =  a2n 
Ex.     (aby)x  =  axbxyx 
Ex.     (2a2b)y  =  2ya2yl>y 


Problems 

1     Define    involution. 

Expand, 

2     (2x2)2 

14 

(ax)2 

3     (4xy) 2 

15 

(2ab) 3 

4     (3*?%) 2 

16 

(2aM)n 

5     (4x2;y)2 

17 

(x2;y)30 

6     (2rry2)3 

18 

(|a26)2 

7     (5^223)4 

19 

(4^)3 

8     ($ab2c)3 

20 

(2a2&*c~2)2 

9     (10c  2d)5 

21 

(4w2w*£-2)3 

10  (2a*>c3)2 

22 

(a*62c~3)2 

11  (3a*)2 

23 

(*Y«f)* 

12  (2**30* 

24 

(2w2w)* 

13  (J)2 

25 

(4a*  6~  V)* 

Lesson  3 

Fractional  exponents. 

1  If  a*    a*    a?  —a5  =a2 

Then  a 2  is  the  product  of  three  equal  factors,  and 
from  the  definition  of  a  root, 
a*  is  the  cube  root  of  a 2 
Or    <y~aT  =  a* 
Stated  in  general  terms,  the  numerator  of  a  frac- 
tional exponent  indicates  the  power  to  which  a  number 
is  to  be  raised,  and  the  denominator  indicates  the  root 
to  be  taken. 

Ex.     a*  means  that  the  4th  root  of  a 3  is  to  be  taken. 
Ex,     x*  means  that  the  3rd  root  of  x2  is  to  be 
taken. 

2  In  the  case  of  a  product  of  two  or  more  factors, 
each  having  a  fractional  exponent,  the  root  can  be 
expressed  as  follows: 

Ex.     a*  b*=ar2b&=y/ a*b3 
Ex.     a*  b*  =  a&x&  =  tya*b3 
Ex.     x^y*  =x£sy^  =  Vx*yl$ 
In  case  an  indicated  root  is  raised  to  a  power. 
Ex.     (^)4  =  (s*)4==xt  =  ^ 
Ex.     Wx)5  =  (yh)5  =  y%  =  Vy_ 
Ex.     Wa)*=(a*)*  =  ai  =  ya~>~ 

3  In  case  a  root  is  to  be  taken  of  a  power. 
Ex.     <s/ai  =  a*  =  \/a 

(142) 


POWERS                                      142 

Ex.     ^a!  = 

3 

9/ 

=  Va2 

Ex      y/Y\  = 

=  &*  = 

-</b 

(The  index  number  becomes  a  factor  of  the  de- 

nominator.) 

Problems 

Write  with  fractional  exponents. 

1     Va 

7     V2a 

2     V^" 

8     \Aw3 

3     V2"3 

9     ^ 

4     #a 

10     ^W 

5    ■#? 

11     VS? 

6     ^2^ 

12  V^7 

13  y~ah 

Write  with  the  radical 

sign. 

14    o* 

18    a*6* 

15     6* 

19     2* 

i6  a 

20    .a*&M 

17     (<*&)* 

21  (akr)f 

22  (*y)= 

Lesson  4 

Zero  and  negative  exponents. 

1  Not  only  do  we  have  integral  and  fractional 
exponents  to  work  with  but  also  the  zero  exponents 
often  appear. 

Any  number  taken  zero  times  as  a  factor  gives 
the  product  one. 
Ex.     2°=1 
Ex.    a°=l 
Ex.     (2abc)°  =  l 

2  The  explanation  of  this  law  is  as  follows : 

From  the  first  law  of  exponents  ax.a°  =  ax+0  =  a* 

ax 
Dividing  the  equation  through  by  ax,  a°  =  —  —  l 

ax 

Another  explanation  is  this, — as  the  unit  (one)  is 
always  used  as  the  base  for  multiplication,  to  use 
a  number  zero  times  as  a  factor  means  that  the  ori- 
ginal (one)  is  not  multiplied  by  anything.  There- 
fore the  result  is  simply  one. 

Ex.     3 -4  =  four  units  taken  as  many  times  as 
one  unit  is  taken  to  get  three. 

32  =  3-3,  indicates  that  3  units  are  taken  twice 
as  a  factor. 

3  Any  number  having  a  negative  exponent  is 
equal  to  one  divided  by  the  number  with  a  positive 
exponent. 

Ex.    a-*=± 


Ex.    x~*=  — 

v  3 


(144) 


POWERS  145 

4  The  proof  of  this  law  follows: 
aT1"-  a+m  =  arm+m  =  a°=l 

a-™-  am=l 

Dividing  through  by  am        a~m  —  ~Hi 

0 

5  This  law  is  made  use  of  in  rewriting  an    ex- 
pression having  a  fractional  form  in  an  integral  form. 

a2 
Ex.     —  =  a2x-3 
x3 

x9y* 

Ex.       _A  0=x3y4w4z~~2 

w     4Z2 

(Note)     Integral  expressions  may  also  be  written 
in  fractional  form  by  reversing  the  process. 


Problems 

Expand 

1     (2a°b) 2 

6     (2ab)° 

2     (3ab°)° 

7  (xyz)° 

3     (O 2 

8     (x23/z)2 

4     (x°b) 3 

9     3a°=? 

5     (2a°b°) 2 

10    \a°b 

Write  with 

positive  exponents. 

11     a~2 

16    x~~  zy~ 4 

12     (ab)~* 

17     m~2n~3 

13     a2b~z 

18     a-*fr-y 

14    a°b~4 

19    afV2 

15     a2b*c~ 

2                     20    x2y~2z~z 

Lesson  5 

Powers  of  binomials 
Special  rules. 
1     It  often  Happens  that  one  wishes  to  raise   a 
binomial  to  some  power,  usually  to  the  second  or 
third,  but  often  to  the  fourth,  fifth,  or  even  higher 
power. 

A  formula  may  be  developed  by  which  all  such 
powers  may  be  found,  but  for  the  present  purposes 
the  forms  of  the  second  and  third  powers  will  be 
memorized. 

As  stated  once  before,  under  the  special  rules  for 
multiplication,  the  rule  for  raising  any  binomial  to 
the  second  power  is, 

Square  the  first  term. 
Add  twice  the  product  of  the  two  terms. 
Add  the  square  of  the  second  term. 
( Note)     This  law  is  based  upon  the  fact  that  every 
multiplication  of  a  binomial  by  itself  gives  such  an 
expression. 

Ex.     (a-\-b)2  =  a2     Square  of  first  term. 
■\-2ab    Twice  the  product  of  two  terms. 
+b2    Square  of  last  term. 
Or,  (a+b)2  =  a2+2ab+b2 
Ex.     (2a  —  b)2  =  4a2    Square  of  first  term. 
+2  (2a)  (  —  b)     Twice  product  of  two  terms. 
+b2    Square  of  last  term. 
Or  (2a-b)2  =  4a2-4ab+b2 
(Note)     It  will  be  noticed  that  twice  the  product 
of  the  two  terms  gives  a  minus  quantity. 

(146) 


POWERS  147 

Ex.     (x  —  4:y)2  =  x2     Square  of  first  term. 

-\-2(x)(  —  4y)     Twice  the  product  of  two  terms . 

-\-16y2     Square  of  the  second  term. 
Or  (x-4:y)2  =  x2-Sxy+16y2 

Ex.     (  —  x—y)2  —  x2     Square  of  first  term. 

-\-2(—x)(  —  y)     Twice    the    product    of    two 
terms. 

+y 2     Square  of  second  term. 
Or  (— x— y)  2  =  x2-\-2xy-\-y2 

2  The  rule  for  raising  a  binomial  to  the  third 
power  is, 

The  cube  of  the  first  term. 

Plus  three  times  the  product  of  the  first  term  squar- 
ed and  the  2nd  term. 

Plus  three  times  the  product  of  the  first  term  and 
the  2nd  term  squared. 

Plus  the  cube  of  the  second  term. 

Ex.     (a+b)3  =  as     The  cube  of  the  first  term. 
+3a2b    Three  times  the  product  of  the  1st 
term  squared  by  the  second  term. 

■\-3ab2     Three  times  the  product  of  the  1st 
term  by  the  second  term  squared. 

+63     The  cube  of  the  second  term. 
Or   (a+b)*  =  a*+3a2b+3ab2+bs 
(Note)     This  rule  is  derived  inductively  from  the 
fact  that  actual  multiplication  in  a  great  number  of 
problems  gives  the  same  results. 

Ex.     (2a  -  b) 3  =  8a 3    The  cube  of  the  1st  term. 
-\-3(2a)2(  —  b)     Three    times    the    product    of 
the  1st  term  squared  by  the  2nd  term. 


148  BRIEF  COURSE  IN  ALGEBRA 

+ 3  (2a)  (-b)2     Three  times  the  product  of  the 
1st  term  by  the  2nd  term  squared. 

+  ( -  b) 3     The  cube  of  the  2nd  term. 
Or    (2a-b)3  =  Sa*-12a2b+6ab2-b* 

Ex.     (-6-2*)  3=  (-6) 3     The  cube  of  the  1st 
term. 

+ 3  ( —  6) 2  ( —  2x)     Three    times    the    product 
of  the  1st  term  squared  by  the  2nd  term. 

+3(  —  6)(  —  2x)2    Three  times  the  product  of 
the  1st  term  by  the  2nd  term  squared. 

+  (-2x)3     The  cube  of  the  2nd  term. 
Or  (-6-2x)  =  -216-216x-72x2-8x3 

3  It  is  found  by  multiplication  that  the  fourth 
power  of  a  binomial  is  equal  to 

(1)  the  fourth  power  of  the  first  term. 

(2)  four  times  the  cube  of  the  first  term  multi- 
plied by  the   second  term. 

(3)  six  times  the  square  of  the  first  term  multi- 
plied by  the  square  of  the  second  term. 

(4)  four  times  the  first  term  multiplied  by  the 
cube  of  the  second  term. 

(5)  the  fourth  power  of  the  second  term. 
Ex.     (a+6)4  =  a4+4a36+6a2&2+4a&3+&4 
Ex.     (x+y)  4  =  x  4+4x  *y+6x  2y  2+4xy  3+y  4 
Ex.     (m+w)  4  =  m4+4w3«+6w2w2-|-4w«8-r-»4 

4  To  overcome  the  difficulty  of  expanding  the 
difference  of  two  numbers  to  any  power,  it  is  best 
to  first  write  a  general  form  then  make  substitutions. 

Ex.     [()  +  0]2  =  ()2+2()()  +  0* 

Ex.    [()  +  ()]3  =  ()3+3()*()+3()()2+()' 


POWERS  149 

Ex.     [()  +  ()] «-()  4+4()  3()+6()  2()  2+4()()  '+ 
04 

If  these  forms  are  studied,  it  will  be  seen  that  the 
exponents  of  the  first  term  decrease  regularly  by  one, 
and  the  exponents  of  the  second  term  increase  regular- 
ly by  one.     The  coefficients  may  be  memorized. 
Suppose  (*— y) 2  is  to  be  expanded. 
Expanding  each  term  and  combining, 
W2+2W(-y)  +  (-y)2 
x2  —  2xy+y2 
If  (a  —  b)2  is  to  be  expanded, 
(a)2+2(a)(-b)  +  (-b)2 
Expanding  each  term  and  combining, 
a2-2ab+b2 


Problems 

Expand 

1  (a+b)2 

13 

(2x-l)3 

2     (a-b)2 

14 

(l-2x)3 

3  (x-y)2 

15 

(a -2b)* 

4  (2x+l)2 

16 

(m+n) 3 

5  (l-3a)2 

17 

(a+b)' 

6  (2x+y)2 

18 

(a-b)' 

7  (2a+362)2 

19 

(x+y)  ' 

8  (l-3ax2)2 

20 

(x-y)' 

9  (a-b)* 

21 

(m+n) 4 

10  (a+b) 3 

22 

(x-1)' 

11  (1-6)3 

23 

(s+1)' 

12  (x+1)3 

24 

(2x-b)' 

25 

(1+x)' 

Lesson  6 

General  rule. 

Any  binomial  may  be  raised  to  any  power  by  ap- 
plication of  a  general  rule  developed  inductively  from 
the  special  cases.  This  rule  is  proved  and  discussed 
in  a  later  discussion.  It  is  here  given  in  working  form 
only. 

Given  (a+b)n 

( Note)  (n)  is  used  as  the  exponent  to  stand  for  a 
general  number. 

1  Write  a  with  exponent  (n)  or  an 

2  The  coefficient  of  the  second  term  is  n. 

The  (a)  in  the  second  term  has  the  exponent 
(n-1). 

The  (b)  in  the  second  term  has  the  exponent  (1) 
The  expansion  thus  far  is  att-\-nan~1b-\ 

3  The  third  term  has  a  coefficient  formed  by 
multiplying  the  coefficient  of  the  second  term  by  the 
exponent  of  (a)  in  the  second  term,  and  dividing  this 
product  by  2.  (a)  takes  the  exponent  (n  —  2),  (b) 
takes  the  exponent  (2). 

The  expansion    thus   far   is   an-\-nan~~lb-\- — - 

an-262_|_ 

4  The  fourth  term  has  a  coefficient  formed  by 
multiplying  the  coefficient  of  the  third  term  by  the 
exponent  of  a  and  dividing  by  (3) .  (a)  takes  the  ex- 
ponent (n  —  3),  (b)  takes  the  exponent  (3) 

(150) 


POWERS  151 

The  expansion  to  this  point  is, 

«    i,  .  n(n—V)     m    9~     n(n—\)  (n— 2)         Q  0 
a"+nan-1b+    v        y  an_2&2+— ^ -  an~3b3 

£  it  .0 

5  The  exponents  for  each  term  decrease  regularly 
with  respect  to  (a),  and  increase  regularly  with  re- 
spect to  (b) 

The  coefficient  of  a  next  term  is  found  from  the  last 
preceding  term  by  multiplying  the  coefficient  by  the 
exponent  of  the  first  term  of  the  binomial,  and  divid- 
ing by  the  number  of  the  term. 

Ex.     (a+6)10 

First  term,  a 1  ° 

Second  term,  10a  9b 

10.9 
Third  term,  —  a8b2 

„        .  10.9.8     fla 

Fourth  term,    ~yr~  a  b3 

etc. 

6  In  case  the  binomial  is  made  up  of  terms  of 
more  than  one  factor  or  terms  of  different  signs,  it 
is  advisable  to  use  a  parentheses  form  in  the  expansion. 

Ex.     (2a  -62)6 

First   developing   the   expansion   of     (()  +  ()) 6, 
we  have, 

0  6+6()  '0+150  40  2+20()  3()  3+15()  2()  4+6() 
05+06 

Filling  the  parentheses, 

(2a)  6+6(2a)  5(~62)  +  15(2a)  4(-&2)  2+20(2a) 3 
(-62)  3+15(2a)  2(-6 2)  4+6(2a)(-62)  5+(-62) 6 


152  BRIEF  COURSE  IN   ALGEBRA 

Expanding, 
64a 6  -  192a  hb  2+240a  *b 4  -  160a  3b  6+60a  2b 8- 
12ab10+b12 

Ex.     (3a2-b)* 

()3+3()2()+3()()H-()3 

(3a2)3+3(3a2)2(-&)  +  3(3a2)(-&)2+(-&)» 
Expanding, 

27a6-27a4o+9a2&2-63 

Any   polynomial    may    be    squared    by    squaring 
each  term,  and  taking  twice  the  product  of  each  term 
by  every  other  term.     The  signs  of  the  terms  deter- 
mine the  signs  of  the  expansion. 
Ex.     (a+b+c) 2 
a2+b2+c2+2ab  +  2ac+2bc 
Ex.     (a-b+c)2 
a2+b2+c2-2ab+2ac-2bc 

This  is  evident  from  the  fact  that  the  multiplica- 
tion of  an  expression  by  itself  allows  two,  and  only 
two,  like  products  of  terms. 
Ex.     a+b+c 
a+b+c 


a' 
+b2 

l+ab+ac 
+ab         +bc 

+ac+bc+c2 

b2+a 

Expand 

1  (a+b) 

2  (a+b) 

2+2ab+2ac+2bc+c2 
Problems 

2                          3     (a+b) 4 
8                          4     (a+b)  ' 

POWERS  153 


5 

(x-y)* 

12 

(*-2)4 

6 

(a-b)* 

13 

(-*-?)* 

7 

(m-\-ri) 3 

14 

(3-2y)» 

8 

(1+*)2 

15 

(2ax+2;y)8 

9 

(2*+3)  ^ 

16 

($abc-2d)1 

10 

(1+3?) « 

17 

(l+2x)3 

11 

(2x+3;y) 6 

18 

(3-2&)3 

ROOTS 
Lesson  1 

Monomials 

1     Rational  roots. 

The  radical  sign  (vO  placed  over  an  expression 
indicates  that  a  root  is  to  be  taken.  The  number 
placed  in  the  (vO  indicates  the  exact  root  to  be  taken. 
This  number  is  called  the  index  number. 


Ex.     \/a3b*    indicates    that   the   third   root   of 
azb*  is  to  be  taken. 

An  indicated  root  that  can  be  exactly  found  is 
called  a  rational  root. 
Ex.     >78  =  2 
Ex.     ^4  =  2 
Ex.     ^/27  =  3 
Ex.     y/a*=a 

( Note)     If  no  index  number  is  given  the  root  to  be 
found  is  the  second. 

2     Irrational   roots. 

An  indicated  root  that  cannot  be  exactly  found  is 
called  an  irrational  root. 

Ex.     ^16,  Vo7;  </&,  etc. 
Indicated  roots  that  cannot  be  exactly  found  can 
only  be  approximated. 

(154) 


ROOTS  155 

An  expression  involving  an  irrational  root  may  be 
simplified  by  one  of  three  methods. 

(a)  When  the  exponent  of  the  quantity  beneath 
the  radical  sign  is  a  factor  of  the  index  number,  the 
index  number  may  be  divided  by  the  exponent. 

Ex.     y~^  =  Va 

Ex.     ^/o7  =  ^a 

Ex.     Vi=-V2>=</2 

(Note)  Often  the  quantity  beneath  the  radical 
sign  may  be  written  in  an  equivalent  form  with  an 
exponent. 

Ex.     ^8  =  ^2T=^2 

Problems 


10 


Simplify 

1  V16  9  vie 

2  V25  10  ^V 

3  vVF  11  </V 

4  Vo7  12  X/oTb* 

5  ^4  13  </(a-by 

6  ^/9  14  ^4? 

7  ^25  15  #81 

8  #49  16  #32 


Lesson  2 

Irrational  Roots — Continued 

(b)  If  the  quantity  beneath  the  radical  sign  is 
factorable,  one  factor  may  be  a  perfect  power.  In 
such  case  its  root  is  written  outside  the  radical  sign. 

Ex.     Vl8,  =  V972  =  V~9:  V2  =  3V2 


Ex.     ^54  =  ^27-2  =  ^27-  ^2=3^2 
Ex.     V2ai  =  V2^  VaT  =  V27  a  =  aV2 
Ex.     ^a^  =  ^a~^  ya  =  aj/~a~ 


1 

V8 

2 

Vl8 

3 

V40 

4 

Vl2 

5 

V20 

6 

V98 

7 

Va7 

8 

Va7 

Problems 

9 

V8a* 

10 

VaW5 

11 

V50 

12 

^54 

13 

^16 

14 

^32 

15 

aJV 

16 

^(a-6)5 

(156) 


Lesson  3 

Irrational  roots 

(c)  If  the  quantity  beneath  the  radical  sign  is  a 
fraction,  the  expression  may  be  simplified  by  multi- 
plying both  members  of  the  fraction  by  such  a  num- 
ber as  will  make  the  denominator  a  perfect  power. 


Ex. 
Ex. 


Vi- 

=  Vi- 

_V3_ 
~V9 

3 

ivs: 

a 

03 

_</a~* 

_Va~> 
a 

-=1  </oT. 

a 

m,     vj =vj -$-^-i  ve: 


Problems 

Simplify 

i  v i 

7   VT 

2     V"| 

8     Vi 

3     VI 

9     Vf 

4     V^ 

6 

10     -v/5 

5     Vi 

11    v^* 

fc2 

6     V* 

12     V^1 

63 

(157) 


Lesson  4 

Reduction  to  same  order 

(d)     Radical  expressions  may  be  reduced  to  the 
same  order  (having  the  same  index  numbers),  by 

1  Writing  the  expressions  with  f rational  exponents. 

2  Writing  the  exponents  as  equivalent  fractions 
having  common  denominators. 

3  Writing  the  expressions  with  the  radical  signs. 
Ex.     Va,  ^a7,  vV 

Writing  with  fractional  exponents,  a?,  a§,  a$ 
Writing    the    exponents    as    equivalent    exponents 
having  a  common  denominator. 

fl  8  9 

ai^,  are,  am 
Writing  the  expressions  in  radical  form. 


Va6, 

Va8, 

%* 

The  expressions  now  have  common  index  n 

Problems 

Reduce 

i  V2;^3" 

5     V^  ^"6" 

2     y/a,</~b 

6     -^m,  Vw,  \/m 

3     ^2,1/3 

7    Ves^e; 

4     V2t</% 

V"2 

8  v^;^ 

(158) 


Lesson    5 

Mixed  numbers  to  entire  surds 

(e)  A  mixed  expression  may  be  written  as  an  en- 
tire surd  by  raising  the  rational  factor  to  the  power 
indicated  by  the  index  number,  and  placing  it  be- 
neath the  radical  sign  as  a  factor. 

Ex.     2V3  =  V4^3  =  Vl2 


Ex.     6y/a  = 

=  V36a 

Ex.     3V67 

=  V9&3 

Problems 

Reduce 

1     3V2 

9 

SbVab 

2     2V5 

10 

^xy\j2a 

3     5V6 

11 

Za\Jb 

4    ay/1) 

12 

2x\/yw 

5     3V10 

13 

6x2\/xy 

6     ay/abc 

14 

3V2 

7    xVy 

15 

2V2 

8     2xVw 

16 

8V6 

(159) 


Lesson  6 

Application  of  the  fundamental  operations  to 
radical  expressions. 

(1)  Radical  expressions  may  be  added  or  subtract- 
ed, provided  they  have  like  radicands  and  like  index 
numbers. 

Ex.     2\/~a+3V~a  =  5Va 

Ex.     6^1rt+4:ysb  =  10y3b 

Ex.     3V~b-2V~b+Wb  =  5V~b 

If  expressions  have  not  like  radical  factors,  it  is 
often  possible  to  reduce  them  to  forms  having  like 
radical  factors. 

Ex.     V2+6VI8-3V8 
Reducing,  2\/2+6-3\/2-3-2V2 
Or    2V2  +  18V2-6V/2  =  14V2 
Ex.     8V27-2V12+4V48 
Reducing,   8-3V3-2-2V3+4-4V3 

Or,    24V3-4V3  +  16V3 
Combining  terms,  36V 3 

(2)  Radical  expressions  may  be  multiplied  or 
divided,  provided  the  index  numbers  are  like. 

Ex.     V3V5  =  Vl5 
Ex.     -\/a-\/b-\/c  =  \/a-b-c- 
Ex.     2V3-4Va-6V36  =  48V9o6 
(160) 


ROOTS  161 

Radical  expressions  may  be  reduced  to  equivalent 
expressions  with  like  index  numbers  by  the  method 
explained  in  the  previous  paragraph. 
Ex.     Va.y~b.J/7 

Writing  with  fractional  exponents  aA.b^.c1** 

Reducing   the   exponents   to   equivalent   fractions 
having  a  common  demoninator. 
Writing  in  radical  form, 

12/— 7    12/77    1S/T 

Multiplying,  y/a*b*c* 

Problems 
Combine 

1  V18+V8+V15 

2  V27  +  V48-\/108 

3  V20-V180  +  V320 


4  Va2b+Vc2b+Vd2b 

5  a/63-V28+\/252 

6  ^16  +  ^54-^128 

7  -V54-V^+Vl50+\/96 


8  ^16a  +  ^81a-^256a 

9  Vs^-VQx'-Vix1 


10  Va2^  +  V4a2^  — Vl6a2^ 
Simplify 

1  Multiply    V3V2-V4 

2  M         ^5V2 

3  "         V7-^2V3 

4  "         ^-^Vl 


162  BRIEF  COURSE  IN  ALGEBRA 


5 

Multiply   ^i-^V^ 

6 

Divide       V3  by  y/2 

7 

V3  by  ^3 

8 

^9  by  V2 

9 

V5  by  ^J 

10 

V28  by  Vl8 

11 

Multiply  \f%  •  \/xy 

12 

Divide  y/2x2y  by  \/xy* 

Lesson  7 

Polynomials 
The  square  root  of  a  polynomial. 

By  applying  the  rule  for  squaring  a  polynomial 
inversely,  it  is  possible  to  test  any  expression  for  a 
square  root  as  follows : 


Ex.     Va*  +4a3  +4a  +  l+6a2 

Arranging  the  terms  in  descending  order  of  ex- 
ponents, 

a4+4a3+6a2+4a+l 

Extracting  the  square  root  of  the  first  term  for  the 
first  term  of  the  quotient, 

a  4-f4a  3+6o  2+4a+ 1  \a*_ 

Squaring  the  quotient  and  subtracting  this  result 
from  the  expression, 

a4+4a3+6a2+4a+l  \a>_ 

n  4 


4a3+6a2+4a+l 


Dividing  the  first  term  of  the  remainder  by  twice 
the  first  term  of  the  quotient, 

<z4+4a3+6a2+4a+l  \a2  +2a 

n  4 


2a2l4a3+6a2+4a+l 

(163) 


164  BRIEF  COURSE  IN  ALGEBRA 

Bringing  this  term  (2a)  down  as  the  second  term  of 
the  trial  divisor,  and  multiplying  by  (2a). 

a4+4a3+6a2+4a+l  \a3 +2a  +  l 

n  4 


2a2+2a|4a3+6a2+4a+l 
4a3+4a2 

2a2+4a  +  l|  2a2+4a+l 
2a2+4a-fl 

(Note)  After  the  remainder  2a2+4a  +  l  has  been 
found,  the  previous  process  is  repeated. 

A     The  quotient  (a2+2a)  is  multiplied  by  2 

B  (2a2)  is  contained  once  in  (2a2).  This  one  is 
placed  in  the  quotient  and  added  to  the  trial  divisor. 

C  The  trial  divisor  is  then  multiplied  by  the  (one) , 
and  subtracted  from  (2a2+4a+l),  the  previous  re- 
mainder. 

Ex.     To  extract  the  square  root  of  4x4-f-4*3  — 
3*2-2x+l 

4x4+4x3-3x2-2x+l  |2*2+*--l 
4x4 


4*2 

+23 

4x3 
4x3 

-3x2- 
+x2 

-2*+l 

4%2 

c-1 

|  -4x2 
-4x2 

-2*+l 
-2*+l 

POLYNOMIALS  165 

Problems 


Test  for  the  square  root. 

1  *4-6x3+llx2-6x+l 

2  4x4+4x3+5x2-f2x+l 

3  a6-2a5+3a4-4a3+3a2-2a+l 

4  4a2-4a6+4a+62-2&-f  1 

5  l  +  2x+3x2+2s3+*4 


Lesson  8 

Arithmetical  numbers 

The  square  root   of  numbers   may  be  taken   as 
follows : 

Ex.     144 
Point  off  from  the  right  in  groups  of  two  digits  each. 

1'44 
Determine  the  nearest  square  root  to  the  number 
in  the  first  group.     It  is  1. 
144  \l_ 
1 


_J  44 
Place  the  square  of  the  quotient  under  the  first 
group  and  subtract.     The  remainder  is  44. 

Multiply  the  quotient  by  two  for  the  first  part  of 
the  trial  divisor. 
1'44[1_ 

_1 

_2J44 
2  is  contained  in  4  twice.     Place  this  number  in 
the  quotient  and  also  in  the  trial  divisor,  then  mul- 
tiply the  trial  divisor  by  it  and  subtract. 
1'44|12 

J 

22)44 
44 

(166) 


POLYNOMIALS  167 

Ex     Take  the  square  root  of  625 
Point  off  6'25 
Determining  the  nearest  square,     6' 25  |2 


225 
Multiply  the  quotient  by  2  for  the  first  trial  divisor. 
6'25  |2_ 


_4J  225 

4  is  contained  5  times  in  22.  Place  5  in  the  quo- 
tient and  in  the  trial  divisor,  then  multiply  by  5  and 
subtract. 

6'25  [25_ 
4 
45 1  225 
225 


Ex.     V25281 
Point  off  2'52'81 
Determine  the  nearest  square  in  the  first  group. 

2'52'81  [1_ 

1 


152 
Multiply  (1)  by  (2)  for  the  first  trial  divisor. 
2'52'81  [1_ 
1 
_2J~152~ 

Although  (2)  is  contained  (7)  times  in  (15),  it  is 
not  possible  to  use  (7),  owing  to  the  fact  that  the  add- 
ing of  (7)  to  the  trial  divisor  will  give  too  large  a 


168 


BRIEF  COURSE  IN  ALGEBRA 


number  when  multiplied  by  (7).     (5)  is  the  largest 
number  we  can  use. 
2'52'81  [15 

1 


J25J152 
125 


2781 
Multiplying  the  quotient  by  (2)  for  the  new  trial 
divisor,  the  process  is  finished  as  follows: 
2'52'81  1 159 
1 

J5J152 
152 


309] 2781 
2781 


(Note)  30  is  contained  in  278  nine  times.  This 
nine  is  placed  in  the  quotient  and  in  the  trial  divisor. 
The  trial  divisor  is  then  multiplied  by  9,  giving  2781. 


Problems 

1 

196 

7 

2304 

2 

484 

8 

3645 

3 

466489 

9 

18496 

4 

8364 

10 

231 

5 

68459 

11 

1002001 

6 

1024 

12 

63945 

Lesson  9 

The   Imaginary  number 

(1)  If  (+a)  and  (+a)  are  multiplied  together  the 
product  is  (+a2) 

If  (—a)  and  (—a)  are  multiplied  together  the  pro- 
duct is  (-\-a2) 

It  is  evident  then  that  the  square  root  of  (+a2) 
is  either  (-{-a)  or  (— a). 

It  often  happens  that  the  even  root  of  a  negative 
number  is  expressed. 
Ex.     xA^i 

Neither  (+a)  nor  (—a)  can  be  used  because  squar- 
ing either  (-f-a)  or  (— a)  will  give  (+4). 

Therefore,  we  define  an  indicated  even  root  of  a 
negative  number  as  an  imaginary  number. 

Because  this  kind  of  a  number  may  appear  in  an 
equation  at  any  time,  it  is  necessary  to  determine  a 
method  of  simplification. 

Given  the  problem  y/— 16 

As  it  stands,  the  square  root  cannot  be  taken, 

But  by  factoring  into  (16)  and  (  —  1),  the  expression 
becomes  Vl6— 1 
Or  VwV^ 

This  equals  4  •  y/~ 1 

It  is  always  possible  to  take  out  the  factor  —  1 ,  so 
that  any  imaginary  number  may  be  thrown  into  the 

(169) 


170  BRIEF  COURSE  IN  ALGEBRA 

form  of  a  radical  multiplied  by  an  imaginary  number. 
The  radical  may  then  be  simplified. 

Ex.     3VII81=3\/8TV:rl  =  3.9V-l  =  27V:::I 

Ex.     6v/:=:49  =  6\/49  V^  =  6.7V-T=42vc:T 

(2)     The   imaginary   number    V— 1    is  called  the 

imaginary  unit.     For  convenience  the  following  table 

T  =  i 
=  -1 

=+1 

(Note)  To  shorten  the  process  of  working  out 
problems  involving  imaginary  numbers,  the  letter  i 
is  used  to  stand  for  V—  1- 

(Note)  After  the  factor  V— T  is  taken  out,  the 
remaining  radical  expression  is  reduced  by  the  laws 
governing  radicals. 


should  be  remembered. 

(V=T)2=-1 

IfV- 

(V-l)3=-V-i 

i2 

(V-l)4=  +  i 

i3 

(V-i)6  =  V-i 

i4 

Problems 

Simplify 

1    v^ 

8 

&Zp. 

2     V-16 

9 

V-50 

3     V^ 

10 

^-44 

4     -^-32 

11 

V-4a62 

5     V-28 

12 

■\/—min 

6     V-a* 

13 

\/-~20 

7    V~x*y* 

14 

^-64 

15 

V11!^ 

Lesson  10 

(3)  Before  applying  the  fundamental  operations  to 
imaginary  numbers,  it  is  necessary  to  write  the  num- 
bers in  equivalent  form  having  the  V— 1  factor  ex- 
pressed. 

Ex.     Add  x/~^A  and  V117!^ 

2V-I+4v/=rT  =  6V-l 
Ex.     Subtract  \J—  a2  from  2\/— 4aT 

(2  •  2a V— T) -a V^  =  4a V^-aV— T  = 

Say/— I 


Ex.     Multiply  V— 4962   by  V-9a26: 
lby/^-3aby/~l  =  2\ab\y/^l)2  = 
21ab2(-l)  =  -21ab2 


Ex.     Divide  V— 25c4  by  V—  c2 
5c  V3!  ■*  c  y/:zl  =  5c 

Often  terms  may  be  added  or  subtracted  even 
though  the  unit  cannot  be  taken  out,  providing  it  is 
possible  to  get  the  even  root  of  some  other  number. 


Ex.     V-2  +  2V-2-V-32 

V-2  +  2  V-2  ~  4  V^  =  -  y/~-2 

(4)     An   expression   involving  an  integer  and  an 
imaginary  number  is  called  a  complex  number. 

Ex.     2  +  v/:i3 

(171) 


172  BRIEF  COURSE   IN  ALGEBRA 

Problems 

Add  Simplify 

1  V^+V1^  8    V-sV^i 

2  V^+V^+V^       9     V^V1^ 

3  V^i-V1^  10    V^V1^ 

4  2vcT6+3V:z25  li    VzT6V-4 

5  V=98-v°72  V^ 

6  3v/-274-4V:l48  V11^ 


7     V-12-V-147  13     V^12 

V-27 


Lesson  11 

Complex  numbers 
Complex  numbers  are  not  difficult  to  handle,  pro- 
viding one  treats  all  operations  involving  the  im- 
aginary terms  with  consideration  for  the  preceding 
rules. 

Ex.    Add  3-rV^2 

This  may  be  written 

4-2V-2 
Adding        7  —  V^-2 


Ex.     Add    6+V-9a2 


2  +  V-16a2 
Re- writing        6+3ay/—  1 
2+4aV-T 


Adding  8+7aV— 1 

Ex.     Multiply     3  -  V— 49 
2  +  Vc:81 


Re- writing  3  —  7  V—  1 

2-f9V:rI 


6-1^ 
-f27V-~l-63(V-I)2 

_6-hl3V:=I-63(V=rl)2 
Or,  6-hl3-s/-I-63(-l) 
Or,  6+13V-1  +  63 

Or,  69  +  13  V11! 

(173) 


174  BRIEF  COURSE  IN   ALGEBRA 

Problems 

Multiply 


Add 

1     2  +  V^3 
3-V-3 


1     5  +  V^ 

2-V-6 


2     4V-8  +  2V-27 

2 

3-V-6 

6V-18-2V-48 

2±y-5 

3     6V-2  +  V-32 

3 

5-^5--2v^S 

4V-2-V-50 

4 

6+4v/:i3 

Subtract 

2  +  3V-2 

1     6-V-3 

5 

6+V-~3 

2  +  V-3 

2-V-4 

2     V^  +  l 

6 

3-V-8 

2V-3-4 

2  +  V-12 

3     6+V-72 

4-2vcr2 

SECTION  V 

THE  QUADRATIC  EQUATION 

Lesson  1 

Equations  in  one  unknown  number. 

(1)  The  equation  studied  thus  far  has  involved 
the  unknown  number  or  numbers  in  the  first  degree 
only,  that  is,  the  exponent  of  the  unknown  symbol 
or  symbols  has  not  been  greater  than  one. 

(2)  The  quadratic  equation  is  one  involving  the 
unknown  numbers  in  the  second  degree,  but  no  higher 
degree.  The  unknown  number  may  appear  in  the 
first  degree. 

Ex.     2x2+3x+6  =  0 
Ex.     ax2+bx+c  =  0 

(3)  To  classify  equations  for  special  solutions  it 
is  customary  to  call  an  equation  involving  the  unknown 
number,  in  the  second  degree  only,  an  incomplete 
quadratic  equation,  and  one  involving  the  unknown 
number  in  both  the  second  and  first  degrees  a  complete 
quadratic  equation. 

Ex .     3x 2  -f-  4  =  8     (Incomplete) 
ax 2 + c  =  0     (Incomplete) 

Ex.     Ax2  —  6x  =  5     (Complete) 

ax2-\-bx+c  =  0     (Complete) 
(175) 


176  BRIEF  COURSE  IN  ALGEBRA 

Solutions 

The  incomplete  quadratic  equation. 

(1)  To  solve  an  incomplete  quadratic  equation 
it  is  only  necessary  to  so  perform  the  fundamental 
operations  upon  the  equation  as  to  get  the  unknown 
number  upon  one  side  of  the  equation  and  the  known 
numbers  upon  the  other  side,  then  extract  the  square 
root  of  both  members. 

Ex.     Given  3**+4  =  8 

Subtracting  4  from  both  members,  3xz  =  4 

Dividing  by  3,  #2  =  y 

Extracting  the  square  root,  x=  ±V% 

Or,   x  =  ±r2Vi 
( Note)     The  ±  sign  is  placed  before  the  number  on 
the  right  hand  side  of  the  equation,  owing  to  the  fact 
that  the  squaring  of  either  -f  V-f  or  —  y/f  will  give  |. 

Ex.     6x2-f2  =  26 
Subtracting  2,  6*2  =  24 
Dividing  by  6,  *2  =  4 
Extracting  the  square  root,  x=±2 

( Note)  The  ±  sign  is  used  as  before  because  either 
+  2  or  —  2  squared  will  give  4. 

Ex.     ax2+b  =  0 
Subtracting  b,  ax2=  —b 
Dividing  by  a,  x2  =  z£ 
Extracting  square  root,  x=ztV~ A 

a 

(Note)  The  values  of  the  unknown  number  are 
called  roots  of  the  equation. 


THE  QUADRATIC  EQUATION 

Problems 

Solve, 

1     x2  =  16 

7     2x2+8  =  33+x2 

2     3x2  =  12 

8     6x2-2  =  22 

3     4x2+2  = 

18 

9     8-*2=-8 

4     ^  =  4 

10     7x2  =  343 

5     ^  =  6 

11     x2+3-2x2  =  7 

177 


6     2x2  — 1=jc2+3  12     -x2+3=-6x2+48 


Lesson  2 
The  complete  quadratic  equation. 

(1)  The  solution  of  the  complete  quadratic  equa- 
tion is  by  the  method  called  completion  of  the  square. 

(2)  It  is  based  upon  the  fact  that  any  trinomial 
having  2  squared  terms  and  the  third  term  equal  to 
twice  the  square  roots  of  the  squared  terms  is  a  perfect 
square. 

Ex.    a*+2ab+b* 
Ex.     x2-f-4x+4    etc. 

(3)  Given  an  equation, 
x*+4x+$  =  0 

It  is  noticed  that  one  term  (x 2)  is  a  square,  so  by 
subtracting  (8)  from  both  members,  x*-\-4:X=  —  8  we 
may  suppose  that  a  third  term  might  be  added  to  the 
left  hand  number,  completing  a  square. 

This  added  term  is  determined  by  supposing  that 
4x  equals  twice  the  product  of  the  square  roots  of  the 
other  two  terms.  Then  division  by  two  would  give  the 
product  of  the  square  roots,  and  again  division  by  one 
square  root  (namely  x)  would  give  the  second  square 
root.     Squaring,  we  have  the  term  to  be  added. 

Example    **+4*+(g)'  =  -*+(£)' 

Simplifying,  #2+4#-f4=—  8-f4 
Or,  s*+4z-f4=-4 
(Note)     This   last    equation   has    now    a    perfect 
square  for  its  left  hand  member. 

(178) 


THE  QUADRATIC  EQUATION  179 

Extracting  the  square  root,  x-\-2  —  ±\/— 4 
Or,    x=±:\/— 4  —  2 

Example     x2  +  5x-(-6  =  0 

Subtracting  6,  x2  +  5x=  —  6 
Completing  square  of  left  hand  member, 

x*  +  5x+\ 


w=-'+(i) 


25  25 

Simplifying,  x2  +  5xJr—  =  —  6  +  — 

n        1A1K     .  25      -24^25 
Or,  x«  +  Sa:  +  T^-T-+T 

Or,  x2+5x+—-  =  - 
4      4 

Extracting  the  square  root  of  both  members, 
,5  IT         1 

2     -\  4      ~2 

Subtracting  -,  x=zb-  — - 

4  6 

Or,  x=—  -=  —  2  and   — r=— 3 

(Note)     The  ±  sign  signifies  that  (J)  is  to  be  used 
first  with  the  (-f-)  sign  then  with  the  (  — )  sign,  giving 
two  values  for  x,  (  —  2  or  —  3). 
Example     x2  —  x  —  6  =  0 
Adding  (6)  to  both  members,  x2  —  x=-\-6 
Completing  square  of  left  hand  member, 


x  + 


Or,   x2-x+]-=+6+)- 
4  4 

n         2         ,1     24,  1     25 

Or,  x2  —  x+-  =  — +-  = — 
4      4      4      4 


180  BRIEF  COURSE  IN  ALGEBRA 

Extracting  square  root  of  both  members, 

1  5 

X  —  _=-4-_ 

2  ~~2 

(Note)     It  will  be  observed  that  the  term  added 

[2nd  term       1  * 


2  •  sq.  rt.  of   1st 


Adding  (J)  to  both  members,  x=±;-+- 

6  4 

Or,  x  =  -  =  3  and  — -=  —2 
2  2 


ProW 

ms 

Solve. 

1     x2+5s+6 

7 

x2-f-8x-f  15 

2     x2+7x+10 

8 

jc2  —  6#— 16 

3     x2— x— 6 

9 

x2+2x-15 

4    s2+13s+30 

10 

w2-w-30 

5     a2+lla+30 

11 

3,  2  +  7y=-  12 

6     &2-26-15 

12 

X2_7:x.=  +18 

13  The  sum  of  the  squares  of  two  consecutive 
numbers  is  120.     Find  the  numbers. 

14  A  lot  has  an  area  of  one  acre.  The  length  is 
2  rods  more  than  the  width.     Find  the  dimensions. 

15  Suppose  a  field  is  80  rods  long  and  20  rods 
wide.  How  wide  a  strip  must  be  cut  to  amount  to 
10  acres. 


Lesson  3 

(4)  In  case  the  term  containing  the  square  of  the 
unknown  number  has  a  coefficient  which  is  not  a 
perfect  square,  it  is  first  necessary  to  either  multiply 
or  divide  by  some  number  to  make  the  coefficient  of 
the  squared  term  a  perfect  square. 

Example    2x2+3*-f4  =  0 

3 
Dividing  by  (2),  x2+-*+2  =  0 

Subtracting  (2),  x2-\— x  =  -2 
Completing  square,  x  2+-x+  I  j-  )   =  ~  2  + 1  7~  I 

Or,  *2-f-s+^=-2+—  =-^ 
2       16  16         16 

Extracting  square  root  of  both  members, 
,3  I     23 

Or,  *m-~±lV^& 

4     4 

Example    3x 2 + 2  lx + 30  =  0 
Dividing  by  (3),  x2+7x+ 10  =  0 
Subtracting  (10),  *2-f7x=-10 

(7*V 
—  I    =  — 10  + 


GO'- 


49 
10+^ 


(181) 


182  BRIEF  COURSE  IN  ALGEBRA 


Extracting  square  root, 

r,                3     7 

Or,  x=  +-  — - 

~~2     2 

,7             9         3 

*  +  -  =  -4-\/  -   =  -4-- 

2     -Af  4      ~2 

n               4         o        ^       10 

Or,  *=--_= -2,  and  —_=-;> 

Problems 

1     2x2-2x=12 

6     3;y2-15;y-42  =  0 

2     3x2+10x  =  12 

7     2x2+3x+2=0 

3     2x2-r-14x=-20 

8     3a2+4a-3  =  0 

4     7a2+7a-42  =  0 

9     2m2+3m-7  =  0 

5     2y2+18y+40  =  0 

10     4x2+3x-2  =  0 

11  A  motor  boat  goes  up  stream  and  back  in  6 
hours.  If  the  boat  goes  up  10  miles  and  the  stream 
has  a  rate  of  4  miles  per  hour,  what  is  the  rate  of  the 
motor  boat  in  still  water. 

12  Sum  of  squares  of  two  numbers  is  73.  Their 
difference  is  5.     Find  the  numbers. 


Lesson  4 

(5)     The  formula. 

The  equation  ax  2±Lbx±1c  =  0  is  called  the  type  form 
of  the  complete  quadratic  equation,  (a)  represents 
the  coefficient  of  the  squared  term,  (b)  represents  the 
coefficient  of  the  first  power  term,  while  (c)  repre- 
sents the  constant  term. 

Example — Given  the  particular  equation, 

3x2+2x+4  =  0 

Then,  a  =  3,  6  =  2,  c  =  4 

If,  then,  we  solve  the  equation, 
ax2+bx+c  =  0, 
we  solve  the  type  of  all  quadratic  equations.     It  is 
then  possible  to  substitute  the  particular  numbers  for 
(a),  (b),  and  (c)  in  the  result  and  get  a  solution  for 
any  particular  equation. 

Example — Given  the  equation 
ax2+bx+c  =  Q 

Solving  this  by  completing  the  square,  we  get 

-&±V&2  — 4ac 

x  = L 

2a 
(Note)     This  solution  is  given  later. 

Now  suppose  the  particular  equation  is 
x2+8x+15  =  0 
ami,  6  =  8,  c  =  15 

(183) 


184  BRIEF  COURSE   IN  ALGEBRA 

Substituting  these  values  of  a,  b,  and  c  in  the  solu- 
tion of  the  type  we  get 


_-&=f=V/62-4ac_-8±:V/64-(4.1.15)  _ 
*  2a  2 


8±V64-60     -8=hv/4      -8±2 


-10        -6  _  , 
or  — —  or  —  5  or  —  3 

2  2 

Example — Given  the  problem  3x2  —  5^+6  =  0 

We  know  that  the  solution  of  the  type  gives 
_  -b±\/b2—  lac 
X~  2a 

In  the  given  problem, 
a  =  3,  b=  —  5,  c  =  6 


Substituting,   x.-(-5)±V(-5)'-(4.3.6) 


e-       rr  :  +5±V25-72 

Simplifying,  x  = 

6 


0r,  g.+5^V-47 

6 
Example — Given  ac2+6^-f8  =  0 
a  =  l,  6  =  6,  c  =  8 


,™         ,                  -6zhV5T-4ac 
Type  form,  x  — 

La 


Substituting,    ,=  -6±V(6)'-(4.1.8) 


THE  QUADRATIC  EQUATION  185 

(Note)  The  given  equation  must  be  arranged  so 
that  all  terms  are  upon  the  left  side  of  the  equality 
sign,  that  is,  arranged  like  the  type  form. 

The  solution  of  the  type  form  follows. 

Problems 
Solve  by  using  the  formula. 


1 

x2+10x+21=0 

7 

3w2+4w+3=0 

2 

2s2+24x+54  =  0 

8 

6yz+2y=-6 

3 

3x2-3x-90  =  0 

9 

5m2+2m-5=0 

4 

5a2-3a+2=0 

10 

3a2-2+3a  =  0 

5 

2y2+3y-4  =  0 

11 

3b2-2b  =  6 

6 

5&2+36+2  =  0 

12 

4c2=-5c+2 

Lesson  5 

Development  of  the  formula 

(6)     Given,  ax2+bx+c  =  0 

Subtracting  (c),  ax2+bx  =  —  c 

Multiplying  both  members  by  (a),  a  2x  2-\-abx  =  —ac 

Completing  square  of  left  hand  member, 

-•+*+(£)'-  -<+(£)' 

b2  b2 

Or,    a2x2+abx+—=-ac+— 

4  4 

Extracting  square  root  of  both  members, 

b 
2 


ax+--=±-y  -ac+-- 


Or,  ax+-  =  ± 


b  l—lac  +  b'' 


0  -  .                b                b         —Aac+b2 
Subtracting  -,  ax=  —  ~±\ r 


2'  2 

b 


Simplifying,   ax  =  —  - ±- V—  4ac+b 


-               —  6±V—  4ac  +  b2 
Or,   a*  = 


-n*-   -j-       u     /  \  —b±z^/b2—4:ac 

Dividing  by  (a),  x  =  ~  - 

Ad 

( Note)  This  solution  may  be  obtained  by  division 
of  both  members  by  the  coefficient  of  (x2),  or  in  fact 
by  multiplying  or  dividing  by  any  number  making 
the  (x 2)  term  a  perfect  square. 

(186) 


THE  QUADRATIC  EQUATION  187 

(7)     If  the  solution  of  the  type  form  of  the  quad- 
ratic equation  gives, 


—  b±<s/b2  — 4ac 
2a 
it  is  evident  that  there  is  in  every  solution  of  par- 
ticular problems  a  value  corresponding  to  b2  —  \ac, 
and  it  is  also  evident  that  it  is  this  value  that  de- 
termines to  a  great  extent  the  value  of  the  quadratic 
equation. 

When  substitutions  are  made,  if  b2  —  Aac  is  positive, 
then  the  quantity  under  the  radical  sign  is  positive, 
and  the  values  of  (x)  are 

real 

unequal  (due  to  dz  sign) . 

When  substitutions  are  made,  if  b 2  —  4ac  is  negative, 
then  the  quantity  under  the  radical  sign  is  negative, 
and  the  values  of  (x)  axe 

imaginary 

unequal  (due  to  ±  sign). 

When  substitutions  are  made,  if  b2  —  4ac  is  zero, 
then  the  quantity  under  the  radical  sign  is  2ero, 
causing  the  radical  term  to  drop  out,  making  the 
values  of  x 

real 

equal  (The  term  preceded  by  the  ±  sign  drops 
out.) 

(8)  This  value,  b2  —  4ac,  is  called  the  discrimi- 
nant, and  by  considering  its  value  it  is  possible  to 
determine  a  great  deal  about  the  values  of  the  equa- 
tion before  the  solution  is  completed. 


188  BRIEF  COURSE   IN  ALGEBRA 

Example    3x2-f-5x+6  =  0 
a  =  3,  6  =  5,  c  —  6, 
62-4ac  =  25-72=-47 
—  47  has  the  (  — )  sign  preceding  it  so  we  know  the 
values  of  (x)  are 
imaginary 
unequal 


~      —  b±^/b2—  4ac_  —  3±V—  47     imaginary 
2a  6  unequal 

Example    2* 2  -f-  4x  —  7  =  0 
a  =  2,  6  =  4,  c=— 7 
6a-4ac  =  16-(4.2.-7) 

=  16-(-56) 

=  16+56 

=  72 
+  72  has  the  (+)  sign  preceding  it,  so  we  know  the 
values  of  (x)  are 
real 
unequal 


Or  ~h±^b2-^ac  =  ~4±V72     real 

2a  4  unequal 

Example    #2-f  4*-t-4  =  0 

a  =  l,  6  =  4,  c  =  4 

62-4ac  =  16-(4.1.4) 

62-4ac=16-16  =  0 

b 2  —  4ac  =  0,  so  we  know  that  the  values  of  (x)  are 

real 

equal 


-6±v/6,-4ac     -4±0 

(jr, -— = *=  —  2  or  —  2 

2a  2 


THE  QUADRATIC  EQUATION  189 

Problems 

Discuss  the  following  problems  with  reference  to 
the  character  of  the  roots. 

1  3x2+2x-3  =  Q  6  y2+3y-6  =  0 

2  6x2-3x+4  =  0  7  2y2+3y-2  =  0 

3  x2-2x+6  =  0  8  3m2+2m-6  =  0 

4  a2-3a+4  =  0  9  6y2+5y-7=0 

5  2a2+2a-6  =  0  10  2r2+3r-f8  =  0 


Lesson  6 

Special  Solution  by  Factoring 

(9)  In  case  the  problem  can  be  arranged  so  that 
the  left  hand  member  is  factorable,  the  solution  is 
easily  accomplished. 

Example    x2  —  x  —  6  =  0 

Factoring  left  hand  member,  (x  —  3)  (x+2)  =  0 

Because   the   product   of   the   two   factors   equals 
zero,  it  is  evident  that  at  least  one  factor  must  equal 
zero.     Inasmuch  as  either  factor  equaling  zero  would 
cause  the  product  to  become  equal  to  zero, 
(x  —  3)  may  equal  zero. 
Or  (x-\-2)  may  equal  zero. 
Ifx-3  =  0 

x  =  3 
If  x+2=0 
x=-2 
Either  of  these  values  for  (x)  will  satisfy  the  equa- 
tion. 

Example    x2-{-5x-\-6  =  0 
Factoring,  (x+3)  (s+2)=0 
If  s+3  =  0,  x=-3 
If  *+2  =  0,  x=-2 
Example    x2+ 7x+10  =  0 
Factoring,  (x+2)  (x+S)  =  0 
If  *4-2  =  0,  x=-2 
If  *+5  =  0,  x=-5 

(190) 


THE  QUADRATIC  EQUATION  191 

Example    x3+6x2+12x-f8  =  0 
Factoring,  (x+2)3  =  0 
If  x+2  =  0,  x=-2 
The  other  two  roots  are  also  equal  to  —  2 . 

(Note)     This    last    equation    is    not    a    quadratic 
equation,  but  may  be  solved  by  factoring. 


Problems 

1 

2x+l=0                           7 

2a  2+34a+ 144  =  0 

2 

*2+10z-f21=0               8 

2(x2-4)=0 

3 

a2+9x+l8  =  0                9 

3x2+21x-f30  =  0 

4 

m2-5m-6  =  0              10 

x2+10x+25  =  0 

5 

x2-r-14*+24  =  0             11 

a2+14a+48  =  0 

6 

5*2+25z+30  =  0           12 

*2-l=0 

Lesson  7 

(10)  A  simple  test  for  the  solution  of  a  quadratic 
equation  is  in  the  fact  that  the  sum  and  the  product 
of  the  roots  have  a  definite  relation  to  the  coefficients 
and  the  constant. 

The  relationship  is, 

Given  the  equation,  ax*-\-bx+c  =  0 

Sum  of  roots  =  —  - 
a 

Product  of  roots  =  - 
a 

Example — Suppose  the  equation,  ^*+8x+15  = 

0  is  solved  giving  the  values,  x=  —3 

Sum  of  roots  =  —  8  =  —  - 
Product  of  roots  =  + 15  = — 

Problems 
Solve  and  test  results. 


1 

x2+17x+30  =  0 

6 

6-3s2+2*  =  0 

2 

x2+27z+50  =  0 

7 

4  —  3x=  —Axr 

3 

a2-10a-39  =  0 

8 

16x2-4:r-2  =  0 

4 

y2-f-10v+24  =  0 

9 

f2+llr+10  =  0 

5 

m2+13w+22  =  0 

10 

2-4*2+3x  =  0 

(192) 


Lesson  8 

(11)     The  proof  of  this  relationship  is  demonstrated 
as  follows: 

Given  the  equation, 

ax2-\-bx-\-c  =  0 
Solving  by  formula, 

-b±Vb2—  A.ac 

x  =  — 

2a 

The  values  of  x  are, 


n\     x_—b  +  y/b2—4ac 
2a 


xl  =  Zb-Vb2-4ac 


Adding,  (x+x1)  = 


2a 

-lb     -b 


2a 


(2)  x-zl+y/*-*™ 

2a  2a 


^-b_Vb2-4ac 


2a  2a 

0i      fri  —  4.ac 

Multiplying,  xxl  = — 

F-y     B  4a2         4a2 

(  Note)     It  will  be  noticed  that  the  two  roots  repre- 
sent the  sum  and  difference  of  the  two  quantities 

—  b      j  b2  —  4ac 

— _  and  — 

2a  2a 

The  product  then  is  the  difference  of  the  squares. 

b2  —  b2-{-4ac     4ac     c 


Combining,  xx*  = 


4a2  4a2 

(193) 


194  BRIEF  COURSE  IN  ALGEBRA 

Problems 

In  the  following  problems  determine  the  sum  and 
product  of  the  roots. 


1 

2x*+3x-4:  =  0 

6 

6a2-f2a-3=0 

2 

x2+4x  -|-4  =  0 

7 

3&H-2&-6  =  0 

3 

3x2-r-2x-4  =  0 

8 

5w2+6  =  3ra 

4 

5a2+6a-2  =  0 

9 

2m2  =  16  —  4m 

5 

3w2-3w  +  2  =  0 

10 

tw  2 
~-3m+6  =  0 

Lesson  9 

Higher  degree  equations 
(12)     Certain    higher    degree    equations    may    be 
solved  by  quadratic  methods,  provided  the  equations 
can  be  placed  in  the  quadratic  form. 
Example    x4+12x2+20  =  0 
Solving  for  (x 2)  instead  of  (x) , 
.     -12±Vl44-80 


2_-12±V64_-12±8_ 
2  2 

-20=-10or^=-2 


2  2  _ 

x=±  V— 10  or  ±  V  —  2 
(Note)     In  such  a  problem  the  unknown  appears 
in  (x2)  and  (x4),  the  one  the  square  of  the  other. 
Example     2x 6  -  2x 3  - 12  =  0 


Dividing  by  2,  re6  —  *3  — 6  = 

0 

Ta_-1±Vl+24 

2 

■     -1±5     -6       +4 
x 3  —              —         or 

2            2           2 

*3=-3  or  2 

Therefore  x  =  <^~3  or  ^2 

^  ,                               Problems 

1     *4+5x2+6  =  0                 6 

y6+12y3+ll=0 

2     2x4-2x2-20  =  0             7 

5x4+2*2-2  =  0 

3     y*+y'-S6  =  0                8 

3m8-2m4=10 

4     6x4  =  24                             9 

6r4+3r2  =  2 

5     a4  +  15a2+26  =  0           10 

2m4  =  32 

(195) 

196  BRIEF  COURSE  IN  ALGEBRA 

Indeterminate  Quadratic  Equations 

1  An  indeterminate  equation  is  satisfied  by  an 
infinite  number  of  sets  of  values  as  was  shown  in  the 
previous  discussion  of  simple  indeterminate  equations. 
The  solutions  discussed  had  to  do  with  the  simultan- 
eous solutions  of  two  or  more  equations,  that  is,  the 
determination  of  a  set  of  values  that  would  satisfy- 
both  or  all  the  equations  under  discussion.  It  was 
found  that  given  simple  equations  could  be  solved 
simultaneously  by  use  of  one  of  three  methods, 
namely,  addition  or  subtraction,  substitution,  or 
comparison.  These  same  methods  may  be  used  when 
second  degree  equations  are  involved.  When  one  or 
both  of  two  equations  are  of  the  second  degree,  it 
is  not  always  possible  to  get  a  solution  by  the  use 
of  one  of  the  three  methods  mentioned,  so  special 
methods  are  employed  for  special  groups  of  equations. 
There  special  methods  will  be  discussed  under  special 
cases  in  the  advanced  course. 

Graphic  Representations 
1  The  graphic  representation  of  an  indeterminate 
quadratic  equation  is  accomplished  by  a  method 
similar  to  that  used  for  the  simple  indeterminate 
equation.  The  fact  that  the  unknown  has  two  values 
makes  possible  the  location  of  two  points  in  each 
set  of  values. 

Ex.     Plot  2*2+;y2  =  4 
Determining  the  sets  of  values, 
lfs  =  0,     y=±2 
If  jc  =  1,     ;y=±V~2~ 
If*  =  2,     y«:fcv^4 


GRAPHIC  REPRESENTATIONS 


197 


Ifx=-1,     y**±V2 
If  x=  —2,     y  =  ±V-4 

If*  =  i,     y=±Vj_ 

Plotting  these  sets  of  values, 


II 


(Note)     Imaginary     values   cannot   be   plotted. 
In  the  graph  it  is  noticed  that  when  x  —  1  there 
are  two  values  of  y  to  locate,  one  above  the  *-axis, 
and  one  below  the  x-axis.     So  with  all  other  values. 
Ex.     Plot,  :y2  =  5s+2 
Determining  sets  of  values, 
lfx  =  0,     y=±x/~2 
If*=l,     y=±vT^ 
If*  =  2,     :y=±Vl2 


198  BRIEF  COURSE  IN  ALGEBRA 

If  x  =  4,     y  =  ±V22 
If**-!,     y**±y/~Z 

Ifx=-2,     y  =  ±y/-K 


■' 

T 

i-* 

\ 

1 

j/ 

H 

x 

J> 

X 

\ 

r1 

r 

-A. 

■ 

X 

,    i 

(Note)  All  equations  of  the  second  degree  have 
the  double  values  to  plot. 

As  in  simple  indeterminate  equations,  two  equa- 
tions may  be  graphically  solved  by  plotting  them  and 
■determining  the  points  of  their  intersection.  These 
points  represent  the  graphical  solution. 


VB  35928 


54  1204 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 


